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course MTH 173
12/15 3:25
Text Section 4.4Text Problem 4.4.1
Given one parameter function f(x) = (x - a)^2
(a)
Let 3 different values of a be 1,2 and 3
The value of the function when a = 1 is
f(x) = (x - 1)^2
The graph of the function will be a concave up graph touching the x axis at x = 1. That is the function y = x^2 is shifted to the right by 1 units. Thus the graph of the function remains positive for all the value of x except x = 1 where the value becomes 0. The y intercept of the graph is (0 - 1)^2 = 1.
The value of the function when a = 2 is
f(x) = (x - 2)^2
The graph of the function will be a concave up graph touching the x axis at x = 2. That is the function y = x^2 is shifted to the right by 2 units. Thus the graph of the function remains positive for all the value of x except x = 2 where the value becomes 0. The y intercept of the graph is (0 - 2)^2 = 4.
The value of the function when a = 3 is
f(x) = (x - 3)^2
The graph of the function will be a concave up graph touching the x axis at x = 3. That is the function y = x^2 is shifted to the right by 3 units. Thus the graph of the function remains positive for all the value of x except x = 3 where the value becomes 0. The y intercept of the graph is (0 - 3)^2 = 9.
(b)
Critical points are the points on the graph where the slope has a value of 0.
When a = 1
The function f(x) = (x - 1)^2
The derivative of the function is f’(x) = 2(x - 1)
The derivative of the function is 0 when 2(x - 1) = 0, that is when x = 1
Thus the critical point of this function occurs when x = 1
The value of the function when x = 1 is 0, thus (1,0) is the critical point of the graph when a = 1.
When a = 2
The function f(x) = (x - 2)^2
The derivative of the function is f’(x) = 2(x - 2)
The derivative of the function is 0 when 2(x - 2) = 0, that is when x = 2
The value of the function when x = 2 is 0, thus (2,0) is the critical point of the graph when a = 2
When a = 3
The function f(x) = (x - 3)^2
The derivative of the function is f’(x) = 3(x - 3)
The derivative of the function is 0 when 3(x - 3) = 0, that is when x = 3
The value of the function when x = 3 is 0, thus (3,0) is the critical point of the graph when a = 3
Thus it can be seen that value of a increases from a = 1 to a = 3 the value the critical points shifts to the right as a increases, the x co-ordinate representing the corresponding a value.
(c)
The formula for x co-ordinate of the critical points in terms of a will be
f(x) = (x - a)^2
Using the rules of differentiation we get
f’(x) = 2(x - a) * 1 = 2(x - a)
to find the critical points we need f’(x) = 0
2(x - a) = 0, x = a
Thus the x co-ordinate of the critical point is a.
Text Problem 4.4.3
Given function f(x) = ax^3 - x
(a)
The 3 different values of a are 1,2,3
When a = 1
f(x) = x^3 - x
The given function is a cubic function with x = +-1 and x = 0 roots. The graph is concave down for x values from - infinity to x = 0 and is concave up from x = 0 to x = +infinity.
When a = 2
f(x) = 2x^3 - x
the given function is a cubic function with x = +- 1/sqrt(2) and x = 0 roots. The graph is concave down for x values from - infinity to x = 0 and is concave up for x values from x = 0 to + infinity. The x coordinate of the critical point will have a value between x = 0 and x = 1/sqrt(2) and x = 0 and x = - 1/sqrt(2).
When a = 3
f(x) 3x^2 - x
the given function is a cubic function with x = +- 1/sqrt(3) and x = 0 roots. The graph is concave down for x values from - infinity to x = 0 and is concave up for x values from x = 0 to x = + infinity. The x coordinate of the critical point will have a value between x = 0 and x = 1/sqrt(3) and x = 0 and x = -1/sqrt(3).
(b)
When a = 1
f(x) = x^3 - x
The derivative of the function is
f’(x) = 3x^2 - 1
For critical points
x^2 = 1/3
x = +- 1/sqrt(3)
When a = 2
f(x) = 2x^3 - x
f’(x) = 6x^2 - 1
For critical points 6x^2 - 1 = 0
x = +- 1/sqrt(6)
When a = 3
f(x) = 3x^3 - x
the derivative of the function is f’(x) = 9x^2 - 1
for critical points f’(x) = 0
x^2 = 1/9
x = +- 1/3
Thus we see that as a increases, the magnitude of the x coordinate of the critical point decreases. The critical points are at equal distance on either side of x = 0 and as x increases they tend to move closer to x = 0.
(c)
Given function f(x) = ax^3 - x
The derivative of the function is
f’(x) = 3ax^2 - 1
The x coordinate of the critical points will be
3ax^2 - 1 = 0
3ax^2 = 1
x = +- 1/sqrt(3a)
Thus the formula for the x coordinate of the critical point is x = +- 1/sqrt(3a)
Text Problem 4.4.6
Given function f(x) = a/x^2 + x, for x > 0.
(a)
The 3 values of a are 1,2 and 3
When a = 1
f(x) = 1/x^2 + x , for x > 0
As x tends to 0 the value of the function tends to infinity and as x tends to infinity the value of the function again tends to infinity.
Also the function is always positive for x > 0, thus the graph will occupy the first quadrant.
The graph is a concave up function, steeply decreasing from infinity, gradually increasing the slope to a smaller negative value finally reaching slope 0 for some x value just greater than x = 1 and then slowly increasing and tending towards infinity.
Thus we have a critical point and also a global minima at x value just greater than x = 1.
When a = 2
f(x) = 2/x^2 + x , for x > 0
As x tends to 0 the value of the function tends to infinity and as x tends to infinity the value of the function again tends to infinity.
The graph of the function is same as that of the function with a = 1, the only difference is in the position if the critical point.
Also the function is always positive for x > 0, thus the graph will occupy the first quadrant.
The graph is a concave up function, steeply decreasing from infinity, gradually increasing the slope to a smaller negative value finally reaching slope 0 for some x value just greater than x = 1.5 and then slowly increasing and tending towards infinity.
Thus we have a critical point and also a global minima at x value just greater than x = 1.5.
When a = 3
f(x) = 3/x^2 + x, for x > 0
As x tends to 0 the value of the function tends to infinity and as x tends to infinity the value of the function again tends to infinity.
The graph of the function is same as that of the function with a = 1 and a = 2, the only difference is in the position if the critical point.
Also the function is always positive for x > 0, thus the graph will occupy the first quadrant.
The graph is a concave up function, steeply decreasing from infinity, gradually increasing the slope to a smaller negative value finally reaching slope 0 for some x value just greater than x = 1.8 and then slowly increasing and tending towards infinity.
Thus we have a critical point and also a global minima at x value just greater than x = 1.8.
Thus we see that the only difference in the 3 graphs is the x value required to reach the minimum value of the function, which is for how long the graph should be decreasing.
(b)
As we can see from the graph that the function moves away from the y axis as value of a increases and thus the x coordinate of the function increases as a increases. Greater the value of a, greater is the value of x which corresponds to the minimum value of the graph that is the critical point.
(c)
Given function f(x) = a/x^2 + x, for x>0
Using the rules of differentiation we get
f’(x) = -2a/x^3 + 1
For the critical points f’(x) should be 0
-2a/x^3 + 1 = 0
1 = 2a/x^3
x^3 = 2a
x = (2a)^(1/3)
Thus the formula for x coordinate of the critical point is x = (2a)^(1/3). This is consistent with the given description of the curve for various values of a.
Text Problem 4.4.8
Given function f(w) = A/w^2 - B/w
Since A and B are positive constants
Using the rule of differentiation, the derivative of the function can be calculated as
f’(w) = -2A/w^3 + B/w^2
For the derivative to be a critical point we need f’(w) = 0
Thus we get B/w^2 = 2A/w^3
We get w = 2A/B
The value of the function when w = 2A/B is
f(2A/B) = AB^2/4A^2 - B^2/2A = B^2( 1/4A - 1/2A) = -B^2/4A
Thus the point (2A/B, -B^2/4A) is a critical point
For the point of inflection we need to find the points where f”(x) = 0
Since f’(w) = -2A/w^3 + B/w^2
The second derivative f”(w) would be = 6A/w^4 - 2B/w^3
For the point of inflection f”(w) = 0
6A/w^4 = 2B/w^3
3A/B = w
The sign of the second derivative when w = 2A/B is positive
The sign of the second derivative when w = 4A/B is negative
Since the sign of the second derivative changes about w = 3A/B, thus w = 3A/B is point of inflection.
The value of the function as w = 3A/B is
f(3A/B) = AB^2/9A^2 - B^2/3A = B^2 ( 1/9A - 1/3A) = -2B^2/9A
Thus the point (3A/B,-2B^2/9A) is an inflection point.
Text Problem 4.4.9
Given function
f(x) = 1 + e^(-ax)
From the graphs it can be clearly seen that the slope at x = 0 is least for graph C, greater for B and maximum for A.
Using the rules of derivative the derivative of the function is
f’(x) = 0 + e^(-ax)*(-a)
The value of the derivative at x = 0 is
f’(x) = 0 - a = -a
The various values of a are a = 1, 2 and 5
The value of the slope at x = 1, 2 and 5 is
f’(1) = -1, f’(2) = -2 and f’(5) = -5
Since the least value if -5, Graph C would correspond to f(x) with a = 5
-2 value is greater than -5 and less than -1, Graph B corresponds to f(x) with a = 2
-1 is the greatest value thus, graph A would correspond to f(x) with a = 1
Text Problem 4.4.12
Given function f(x) = x^3 - ax^2 + b
(a)
Given that b = 1
Thus f(x) = x^3 - ax^2 + 1
I assume 3 values of a as 1,2 and 3
When a = 1
f(x) = x^3 - x^2 + 1
The graph of this function has only 1 real root and other 2 imaginary roots. The real root is a negative root. The graph is concave down for x values from - infinity to some positive value and then concave up for remaining values.
When a = 2
f(x) = x^3 - 2x^2 + 1
The graph of this function has 3 real roots, 1 of which is negative and other 2 are positive. The graph is concave down for x values from - infinity to a x value around x = 0.5 and concave up for remaining values of x.
When a = 3
f(x) = x^3 - 3x^2 + 1
The graph of this function is similar to that when a = 2. That is the function has 3 real roots, one negative and 2 positive. The graph is concave down for x values from - infinity to some x values near x = 1. The graph is concave up for remaining x values. the local maximum and minimum value of the function is less than the value of the function with a = 2.
(b)
Given that a = 1 remains constant.
I assume 3 values of b as b = 1, 2, and 3
When b = 1
The value of the function f(x) = x^3 - x^2 + 1
The graph of the function has only one negative real root and other imaginary roots. The y intercept of this graph is 0 + 0 + 1 = 1. The concavity of the function changes at an x value around x = 0.25.
When b = 2
The value of the function f(x) = x^3 - x^2 + 2
Since the coefficient of x^3 and x^2 remains unchanged, the shape of the graph of the function does not change and remains the same. The y intercept of the graph of the function is 2. Thus we see that the graph of the function only shift upward by 1 unit as compared to the graph of the function with b = 1
When b = 3
The value of the function f(x) = x^3 - x^2 + 3
Similarly, the graph of the function just shifts upward by 1 unit as compared to the graph with b = 2. The shape of the function remains unchanged. Thus the x co-ordinate of the critical point and the inflection point does not change. Only the y value corresponding to that x value shifts by 1 unit as compared to the previous b values
(c)
As b varies we have seen that the graph just shifts upward. Thus for the critical point as well the point will only shift upward as value of b increases. The x coordinate of the critical point remains unchanged, and the y coordinate increases as value of b increases. The difference between 2 consecutive critical points for different values of b will be equal to the difference in the values of b.
As a varies we see that the function moves towards x axis, that is it reduces and stretches, and thus the x coordinate of the critical point increases. Whereas, a critical point with x co-ordinate 0 remains same for all values of a. For the second critical point as a increases, the x coordinate of the critical point increases. The value of the function as this critical point depends upon a as well as b. If b remains constant, the value of the critical point decreases as a increases.
(d)
Given function f(x) = x^3 - ax^2 + b
To find the critical points we need to find the derivative of the function
The derivative of the function will be
f’(x) = 3x^2 - 2ax + 0
To find the critical point, this derivative should be 0
Thus 3x^2 = 2ax
x = 0, or x = 2a/3
Thus the formula for the critical points of the function is x = 0 and x = 2a/3
The value of x indicates that it does not depend upon b and the critical point x = 2a/3 increases as a increases.
Text Problem 4.4.15
Given function f(x) = sqrt(b - (x - a)^2)
(a)
Given that b remains constant and has a value of 1
f(x) = sqrt(1 - (x - a)^2)
3 different values of a assumed are 1, 2 and 3
The value of f(x) when a = 1
f(x) = sqrt( 1 - x^2 + 2x - 1) = sqrt(2x - x^2)
Since sqrt function should have only nonnegative inputs and nonnegative outputs, the domain of this function becomes [0,2] within this domain the graph of the function is concave down almost semicircle starting from origin to the point (2,0). Thus the graph will have only one critical point, the x coordinate of which will have a value around x = 1.
When a = 2
The value of the function f(x) when a = 2
f(x) = sqrt( 1 - x^2 + 4x - 4) = sqrt(4x - x^2 - 3)
Similar to the previous graph, neglecting the negative inputs the domain of the function is [1,3]. The graph of the function is again concave down similar to a semicircle starting from the point [1,0] and moving towards [3,0]. The graph of the function has only one critical point, the x coordinate of which will always have a value around x = 2. The value of the function will always have non negative outputs. This indicate that the graph of the function f(x) when a = 2 is the graph of the function shifted by 1 unit on the right hand side along x axis as compared to the graph of the function when a = 1.
The value of the function f(x) when a - 3
f(x) = sqrt( 1 - x^2 + 6x - 9) = sqrt(6x - x^2 - 8)
the graph of this function is same as the graph of the function when a = 2 shifted to the right by 1 unit. Thus the domain of this function [2,4].
(b)
When a = 1
The function becomes f(x) = sqrt(b - (x - 1)^2)
We assume 3 valeus of b as 1,2 and 3
When b = 1
f(x) = sqrt(1 - (x - 1)^2) = sqrt(2x - x^2)
The graph of the function is almost semicircle that is concave down, starting from origin and ending to (2,0). The critical point of the function has an x coordinate equal to 1.
When b = 2
f(x) = sqrt(2 - x^2 - 1 + 2x) = sqrt(2x - x^2 + 1)
The given function has a domain of [1 - sqrt(2) , 1 + sqrt(2)]
The midpoint or the critical point still has a x coordinate of 1. Thus as compared to the previous semicircle, this semicircle is just stretched, keeping all the other factors same. Distance between the 2 end points ( or the points at which the value of the function is 0) is 2 for b = 1 and now when b = 2 the distance increases to 2sqrt(2)
When b = 3
f(x) = sqrt(3 - x^2 + 2x - 1) = sqrt(2x - x^2 + 2)
The given function has a domain of [1 - sqrt(3), 1 + sqrt(3)]
The midpoint or the critical point still has a x coordinate of 1. Thus as compared to the previous 2 semicircle with b = 1 and b = 2, this semicircle is yet more stretched, keeping all the other factors same. Distance between the 2 endpoints here increases to 2 sqrt(3).
The value of the function always remains positive for all b values when a = 1. The graph remains a semicircle which is concave down with center (1,0) and increasing span(radius) as value of b increases.
(c)
We see that as a increases the graph of the function shifts to the right as a increases and thus the x coordinate of the critical point increases as a increases. The y value of the critical point remains unchanged for any a value.
As value of b increases the radius of the semicircle increases, the center remaining (1,0). Thus we see that the critical point of the function only increases along the y axis. As b increases, the value of the critical point increases, the point does not shift to the right or the left.
Thus we see that increasing a increased the x coordinate of the critical point and increasing b increases the y coordinate of the critical point correspondingly.
(d)
Given function f(x) = sqrt(b + (x - a)^2)
Using the rules of differentiation we get the derivative of the function as
f’(x) = (sqrt(b + (x - a)^2))’ = 1/(2sqrt(b + (x - a)^2)) * ((b + (x - a)^2))’
f’(x) = 1/(2sqrt(b + (x - a)^2)) * (0 + 2(x - a)*(1))
f’(x) = (2(x - a))/(2sqrt(b + (x - a)^2))
For critical point f’(x) should be 0
Since (2sqrt(b + (x - a)^2) is always positive
2(x - a) = 0
x = a
Thus the equation for the x coordinate of the critical points is x = a.
This proves the consistency that the x coordinate of critical point increases as value of a increases and is unaffected by the value of b.
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#$&*
course MTH 173
12/15 3:25
Text Section 4.4Text Problem 4.4.1
Given one parameter function f(x) = (x - a)^2
(a)
Let 3 different values of a be 1,2 and 3
The value of the function when a = 1 is
f(x) = (x - 1)^2
The graph of the function will be a concave up graph touching the x axis at x = 1. That is the function y = x^2 is shifted to the right by 1 units. Thus the graph of the function remains positive for all the value of x except x = 1 where the value becomes 0. The y intercept of the graph is (0 - 1)^2 = 1.
The value of the function when a = 2 is
f(x) = (x - 2)^2
The graph of the function will be a concave up graph touching the x axis at x = 2. That is the function y = x^2 is shifted to the right by 2 units. Thus the graph of the function remains positive for all the value of x except x = 2 where the value becomes 0. The y intercept of the graph is (0 - 2)^2 = 4.
The value of the function when a = 3 is
f(x) = (x - 3)^2
The graph of the function will be a concave up graph touching the x axis at x = 3. That is the function y = x^2 is shifted to the right by 3 units. Thus the graph of the function remains positive for all the value of x except x = 3 where the value becomes 0. The y intercept of the graph is (0 - 3)^2 = 9.
(b)
Critical points are the points on the graph where the slope has a value of 0.
When a = 1
The function f(x) = (x - 1)^2
The derivative of the function is f’(x) = 2(x - 1)
The derivative of the function is 0 when 2(x - 1) = 0, that is when x = 1
Thus the critical point of this function occurs when x = 1
The value of the function when x = 1 is 0, thus (1,0) is the critical point of the graph when a = 1.
When a = 2
The function f(x) = (x - 2)^2
The derivative of the function is f’(x) = 2(x - 2)
The derivative of the function is 0 when 2(x - 2) = 0, that is when x = 2
The value of the function when x = 2 is 0, thus (2,0) is the critical point of the graph when a = 2
When a = 3
The function f(x) = (x - 3)^2
The derivative of the function is f’(x) = 3(x - 3)
The derivative of the function is 0 when 3(x - 3) = 0, that is when x = 3
The value of the function when x = 3 is 0, thus (3,0) is the critical point of the graph when a = 3
Thus it can be seen that value of a increases from a = 1 to a = 3 the value the critical points shifts to the right as a increases, the x co-ordinate representing the corresponding a value.
(c)
The formula for x co-ordinate of the critical points in terms of a will be
f(x) = (x - a)^2
Using the rules of differentiation we get
f’(x) = 2(x - a) * 1 = 2(x - a)
to find the critical points we need f’(x) = 0
2(x - a) = 0, x = a
Thus the x co-ordinate of the critical point is a.
Text Problem 4.4.3
Given function f(x) = ax^3 - x
(a)
The 3 different values of a are 1,2,3
When a = 1
f(x) = x^3 - x
The given function is a cubic function with x = +-1 and x = 0 roots. The graph is concave down for x values from - infinity to x = 0 and is concave up from x = 0 to x = +infinity.
When a = 2
f(x) = 2x^3 - x
the given function is a cubic function with x = +- 1/sqrt(2) and x = 0 roots. The graph is concave down for x values from - infinity to x = 0 and is concave up for x values from x = 0 to + infinity. The x coordinate of the critical point will have a value between x = 0 and x = 1/sqrt(2) and x = 0 and x = - 1/sqrt(2).
When a = 3
f(x) 3x^2 - x
the given function is a cubic function with x = +- 1/sqrt(3) and x = 0 roots. The graph is concave down for x values from - infinity to x = 0 and is concave up for x values from x = 0 to x = + infinity. The x coordinate of the critical point will have a value between x = 0 and x = 1/sqrt(3) and x = 0 and x = -1/sqrt(3).
(b)
When a = 1
f(x) = x^3 - x
The derivative of the function is
f’(x) = 3x^2 - 1
For critical points
x^2 = 1/3
x = +- 1/sqrt(3)
When a = 2
f(x) = 2x^3 - x
f’(x) = 6x^2 - 1
For critical points 6x^2 - 1 = 0
x = +- 1/sqrt(6)
When a = 3
f(x) = 3x^3 - x
the derivative of the function is f’(x) = 9x^2 - 1
for critical points f’(x) = 0
x^2 = 1/9
x = +- 1/3
Thus we see that as a increases, the magnitude of the x coordinate of the critical point decreases. The critical points are at equal distance on either side of x = 0 and as x increases they tend to move closer to x = 0.
(c)
Given function f(x) = ax^3 - x
The derivative of the function is
f’(x) = 3ax^2 - 1
The x coordinate of the critical points will be
3ax^2 - 1 = 0
3ax^2 = 1
x = +- 1/sqrt(3a)
Thus the formula for the x coordinate of the critical point is x = +- 1/sqrt(3a)
Text Problem 4.4.6
Given function f(x) = a/x^2 + x, for x > 0.
(a)
The 3 values of a are 1,2 and 3
When a = 1
f(x) = 1/x^2 + x , for x > 0
As x tends to 0 the value of the function tends to infinity and as x tends to infinity the value of the function again tends to infinity.
Also the function is always positive for x > 0, thus the graph will occupy the first quadrant.
The graph is a concave up function, steeply decreasing from infinity, gradually increasing the slope to a smaller negative value finally reaching slope 0 for some x value just greater than x = 1 and then slowly increasing and tending towards infinity.
Thus we have a critical point and also a global minima at x value just greater than x = 1.
When a = 2
f(x) = 2/x^2 + x , for x > 0
As x tends to 0 the value of the function tends to infinity and as x tends to infinity the value of the function again tends to infinity.
The graph of the function is same as that of the function with a = 1, the only difference is in the position if the critical point.
Also the function is always positive for x > 0, thus the graph will occupy the first quadrant.
The graph is a concave up function, steeply decreasing from infinity, gradually increasing the slope to a smaller negative value finally reaching slope 0 for some x value just greater than x = 1.5 and then slowly increasing and tending towards infinity.
Thus we have a critical point and also a global minima at x value just greater than x = 1.5.
When a = 3
f(x) = 3/x^2 + x, for x > 0
As x tends to 0 the value of the function tends to infinity and as x tends to infinity the value of the function again tends to infinity.
The graph of the function is same as that of the function with a = 1 and a = 2, the only difference is in the position if the critical point.
Also the function is always positive for x > 0, thus the graph will occupy the first quadrant.
The graph is a concave up function, steeply decreasing from infinity, gradually increasing the slope to a smaller negative value finally reaching slope 0 for some x value just greater than x = 1.8 and then slowly increasing and tending towards infinity.
Thus we have a critical point and also a global minima at x value just greater than x = 1.8.
Thus we see that the only difference in the 3 graphs is the x value required to reach the minimum value of the function, which is for how long the graph should be decreasing.
(b)
As we can see from the graph that the function moves away from the y axis as value of a increases and thus the x coordinate of the function increases as a increases. Greater the value of a, greater is the value of x which corresponds to the minimum value of the graph that is the critical point.
(c)
Given function f(x) = a/x^2 + x, for x>0
Using the rules of differentiation we get
f’(x) = -2a/x^3 + 1
For the critical points f’(x) should be 0
-2a/x^3 + 1 = 0
1 = 2a/x^3
x^3 = 2a
x = (2a)^(1/3)
Thus the formula for x coordinate of the critical point is x = (2a)^(1/3). This is consistent with the given description of the curve for various values of a.
Text Problem 4.4.8
Given function f(w) = A/w^2 - B/w
Since A and B are positive constants
Using the rule of differentiation, the derivative of the function can be calculated as
f’(w) = -2A/w^3 + B/w^2
For the derivative to be a critical point we need f’(w) = 0
Thus we get B/w^2 = 2A/w^3
We get w = 2A/B
The value of the function when w = 2A/B is
f(2A/B) = AB^2/4A^2 - B^2/2A = B^2( 1/4A - 1/2A) = -B^2/4A
Thus the point (2A/B, -B^2/4A) is a critical point
For the point of inflection we need to find the points where f”(x) = 0
Since f’(w) = -2A/w^3 + B/w^2
The second derivative f”(w) would be = 6A/w^4 - 2B/w^3
For the point of inflection f”(w) = 0
6A/w^4 = 2B/w^3
3A/B = w
The sign of the second derivative when w = 2A/B is positive
The sign of the second derivative when w = 4A/B is negative
Since the sign of the second derivative changes about w = 3A/B, thus w = 3A/B is point of inflection.
The value of the function as w = 3A/B is
f(3A/B) = AB^2/9A^2 - B^2/3A = B^2 ( 1/9A - 1/3A) = -2B^2/9A
Thus the point (3A/B,-2B^2/9A) is an inflection point.
Text Problem 4.4.9
Given function
f(x) = 1 + e^(-ax)
From the graphs it can be clearly seen that the slope at x = 0 is least for graph C, greater for B and maximum for A.
Using the rules of derivative the derivative of the function is
f’(x) = 0 + e^(-ax)*(-a)
The value of the derivative at x = 0 is
f’(x) = 0 - a = -a
The various values of a are a = 1, 2 and 5
The value of the slope at x = 1, 2 and 5 is
f’(1) = -1, f’(2) = -2 and f’(5) = -5
Since the least value if -5, Graph C would correspond to f(x) with a = 5
-2 value is greater than -5 and less than -1, Graph B corresponds to f(x) with a = 2
-1 is the greatest value thus, graph A would correspond to f(x) with a = 1
Text Problem 4.4.12
Given function f(x) = x^3 - ax^2 + b
(a)
Given that b = 1
Thus f(x) = x^3 - ax^2 + 1
I assume 3 values of a as 1,2 and 3
When a = 1
f(x) = x^3 - x^2 + 1
The graph of this function has only 1 real root and other 2 imaginary roots. The real root is a negative root. The graph is concave down for x values from - infinity to some positive value and then concave up for remaining values.
When a = 2
f(x) = x^3 - 2x^2 + 1
The graph of this function has 3 real roots, 1 of which is negative and other 2 are positive. The graph is concave down for x values from - infinity to a x value around x = 0.5 and concave up for remaining values of x.
When a = 3
f(x) = x^3 - 3x^2 + 1
The graph of this function is similar to that when a = 2. That is the function has 3 real roots, one negative and 2 positive. The graph is concave down for x values from - infinity to some x values near x = 1. The graph is concave up for remaining x values. the local maximum and minimum value of the function is less than the value of the function with a = 2.
(b)
Given that a = 1 remains constant.
I assume 3 values of b as b = 1, 2, and 3
When b = 1
The value of the function f(x) = x^3 - x^2 + 1
The graph of the function has only one negative real root and other imaginary roots. The y intercept of this graph is 0 + 0 + 1 = 1. The concavity of the function changes at an x value around x = 0.25.
When b = 2
The value of the function f(x) = x^3 - x^2 + 2
Since the coefficient of x^3 and x^2 remains unchanged, the shape of the graph of the function does not change and remains the same. The y intercept of the graph of the function is 2. Thus we see that the graph of the function only shift upward by 1 unit as compared to the graph of the function with b = 1
When b = 3
The value of the function f(x) = x^3 - x^2 + 3
Similarly, the graph of the function just shifts upward by 1 unit as compared to the graph with b = 2. The shape of the function remains unchanged. Thus the x co-ordinate of the critical point and the inflection point does not change. Only the y value corresponding to that x value shifts by 1 unit as compared to the previous b values
(c)
As b varies we have seen that the graph just shifts upward. Thus for the critical point as well the point will only shift upward as value of b increases. The x coordinate of the critical point remains unchanged, and the y coordinate increases as value of b increases. The difference between 2 consecutive critical points for different values of b will be equal to the difference in the values of b.
As a varies we see that the function moves towards x axis, that is it reduces and stretches, and thus the x coordinate of the critical point increases. Whereas, a critical point with x co-ordinate 0 remains same for all values of a. For the second critical point as a increases, the x coordinate of the critical point increases. The value of the function as this critical point depends upon a as well as b. If b remains constant, the value of the critical point decreases as a increases.
(d)
Given function f(x) = x^3 - ax^2 + b
To find the critical points we need to find the derivative of the function
The derivative of the function will be
f’(x) = 3x^2 - 2ax + 0
To find the critical point, this derivative should be 0
Thus 3x^2 = 2ax
x = 0, or x = 2a/3
Thus the formula for the critical points of the function is x = 0 and x = 2a/3
The value of x indicates that it does not depend upon b and the critical point x = 2a/3 increases as a increases.
Text Problem 4.4.15
Given function f(x) = sqrt(b - (x - a)^2)
(a)
Given that b remains constant and has a value of 1
f(x) = sqrt(1 - (x - a)^2)
3 different values of a assumed are 1, 2 and 3
The value of f(x) when a = 1
f(x) = sqrt( 1 - x^2 + 2x - 1) = sqrt(2x - x^2)
Since sqrt function should have only nonnegative inputs and nonnegative outputs, the domain of this function becomes [0,2] within this domain the graph of the function is concave down almost semicircle starting from origin to the point (2,0). Thus the graph will have only one critical point, the x coordinate of which will have a value around x = 1.
When a = 2
The value of the function f(x) when a = 2
f(x) = sqrt( 1 - x^2 + 4x - 4) = sqrt(4x - x^2 - 3)
Similar to the previous graph, neglecting the negative inputs the domain of the function is [1,3]. The graph of the function is again concave down similar to a semicircle starting from the point [1,0] and moving towards [3,0]. The graph of the function has only one critical point, the x coordinate of which will always have a value around x = 2. The value of the function will always have non negative outputs. This indicate that the graph of the function f(x) when a = 2 is the graph of the function shifted by 1 unit on the right hand side along x axis as compared to the graph of the function when a = 1.
The value of the function f(x) when a - 3
f(x) = sqrt( 1 - x^2 + 6x - 9) = sqrt(6x - x^2 - 8)
the graph of this function is same as the graph of the function when a = 2 shifted to the right by 1 unit. Thus the domain of this function [2,4].
(b)
When a = 1
The function becomes f(x) = sqrt(b - (x - 1)^2)
We assume 3 valeus of b as 1,2 and 3
When b = 1
f(x) = sqrt(1 - (x - 1)^2) = sqrt(2x - x^2)
The graph of the function is almost semicircle that is concave down, starting from origin and ending to (2,0). The critical point of the function has an x coordinate equal to 1.
When b = 2
f(x) = sqrt(2 - x^2 - 1 + 2x) = sqrt(2x - x^2 + 1)
The given function has a domain of [1 - sqrt(2) , 1 + sqrt(2)]
The midpoint or the critical point still has a x coordinate of 1. Thus as compared to the previous semicircle, this semicircle is just stretched, keeping all the other factors same. Distance between the 2 end points ( or the points at which the value of the function is 0) is 2 for b = 1 and now when b = 2 the distance increases to 2sqrt(2)
When b = 3
f(x) = sqrt(3 - x^2 + 2x - 1) = sqrt(2x - x^2 + 2)
The given function has a domain of [1 - sqrt(3), 1 + sqrt(3)]
The midpoint or the critical point still has a x coordinate of 1. Thus as compared to the previous 2 semicircle with b = 1 and b = 2, this semicircle is yet more stretched, keeping all the other factors same. Distance between the 2 endpoints here increases to 2 sqrt(3).
The value of the function always remains positive for all b values when a = 1. The graph remains a semicircle which is concave down with center (1,0) and increasing span(radius) as value of b increases.
(c)
We see that as a increases the graph of the function shifts to the right as a increases and thus the x coordinate of the critical point increases as a increases. The y value of the critical point remains unchanged for any a value.
As value of b increases the radius of the semicircle increases, the center remaining (1,0). Thus we see that the critical point of the function only increases along the y axis. As b increases, the value of the critical point increases, the point does not shift to the right or the left.
Thus we see that increasing a increased the x coordinate of the critical point and increasing b increases the y coordinate of the critical point correspondingly.
(d)
Given function f(x) = sqrt(b + (x - a)^2)
Using the rules of differentiation we get the derivative of the function as
f’(x) = (sqrt(b + (x - a)^2))’ = 1/(2sqrt(b + (x - a)^2)) * ((b + (x - a)^2))’
f’(x) = 1/(2sqrt(b + (x - a)^2)) * (0 + 2(x - a)*(1))
f’(x) = (2(x - a))/(2sqrt(b + (x - a)^2))
For critical point f’(x) should be 0
Since (2sqrt(b + (x - a)^2) is always positive
2(x - a) = 0
x = a
Thus the equation for the x coordinate of the critical points is x = a.
This proves the consistency that the x coordinate of critical point increases as value of a increases and is unaffected by the value of b.
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