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course MTH 173
10/16 00:55
Text_10Problem 1.6.9
Given function y = 3 * x^-4
The given function is rearranged as y = 3 / x^4
Here the given function is an even function, thus the function is symmetric about y axis.
We see that as the value of x increases the value x^4 will increase and thus the denominator increases, and as the denominator increases the fraction 1 / x^4 decreases and thus the overall value decreases continuously, thus we can see from the graph that as the value of the function increases, that as x tends to + infinity the value of the function tends to 0 from the positive y axis.
Since the function is an even function, thus the function will behave the same way it for the negative values of x as it behaves for positive values of x. Thus the function will tend to 0 from the positive y axis even when x tends to - infinity.
The given function completely resides in the positive y region and the line y = 0 is an asymptote to the function.
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More specifically the positive x axis is an asymptote.
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Problem 1.6.12
Given functions 10*(e^0.1x) and 5000*(x^2)
An exponential function always grows at a faster rate than a power function, that is an exponential function always dominates the power function in the long run, provided that the base of the exponential function is greater than 1, which is the case shown here. Thus when x tends to infinity, the exponential function will be above the power function and thus 10*(e^.1x) dominates 5000*(x^2).
Problem 1.6.14
Given functions 2x^4 and 10x^3 + 25x^2 + 50x + 100
Let f(x) be the function of 2x^4 and g(x) = 10x^3 + 25x^2 + 50x + 100
The derivative that is the rate at which the function changes
f’(x) = 8x^3 and g’(x) = 30x^2 + 50x + 50
f’(10) = 8000 and g’(10) = 3550, and since the power of the f’(x) is greater than g’(x) thus the value of f’(x) will increase at a greater rate than g’(x). and since we see that at just x = 10 f’(x) is greater than g’(x) , f’(x) will remain greater than g’(x) for all x beyond that. Now if the slope, that is the rate of change of the function f(x) increases at an rapid rate as compared to g(x) function, thus the function f(x) would also increase at a much rapid rate as compared to g(x) an in the long run would dominate g(x). Thus when x tends to infinity 2x^4 dominates 10x^3 + 25x^2 + 50x + 100.
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Good.
An alternative:
A long division of f(x) / g(x) would yield
x/5 - 1/2 + ( 1/2 x^2 + x + 10) / (2x^3 + 5 x^2 + 10 x + 20)
The last term approaches zero as x approaches infinity. The x/5 approaches infinity, -1/2 is constant, so the quotient approaches infinity.
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Problem 1.6.15
Given functions 20x^4 + 100x^2 + 5x and 3x^5 + x^3 - 40x^2 + 25.
Let f(x) = 20x^4 + 100x^2 + 5x and g(x) = 3x^5 + x^3 - 40 x^2 + 25
Thus we have f’(x) = 80x^3 + 200x^2 + 5 and g’(x) = 15x^4 + 3x^2 - 80x
Since we have f’(x) to be a cubic function and g’(x) to be a quartic function for larger values of x the quartic function will result in higher values and thus the slope will be greater than the slope of the f(x) function. Thus if slope will be high the value too will be high as the highest power for both the function has positive coefficient, thus the values will be positive and will be quite high. Thus when x tends to infinity g(x) will dominate f(x).
Problem 1.6.19
Given roots of the cubic function ( 2 , 0 ) and ( -2 , 0 ). We also know that for any given polynomial function, whenever a function cuts the x axis, odd number of roots are common with the point of intersection and when it touched even number of root are common. Since we know that the function is cubic, and the function cuts the x axis at ( 2 , 0 ), thus one root is common and since at the point ( -2 , 0 ) touches the x axis, thus 2 roots are common at that point, thus the roots are ( -2 , 0 ) and ( -2 , 0 ). Thus for the cubic function the roots are (x - 2), ( x + 2 ) and (x + 2).
Thus the function y = k *( x - 2 ) ( x + 2 ) * ( x + 2 ) , in order to find the value of k it is given to us that the intercept is 4, that is the function passes through ( 0 , 4)
Thus 4 = k * ( -2 ) ( 2 ) ( 2 )
k = - 1/2. Thus y = (-1/2)*(x - 2)(x + 2)(x + 2) = (-1/2)*(x^2 - 4)*(x + 2) = (-1/2)*(x^3 + 2x^2 - 4x - 8)
Thus the cubic function is y = -x^3/2 - x^2 + 2x + 4
Problem 1.6.20
Given function cuts the x axis at 3 points, thus the simplest function would be a cubic function. The given function cuts the x axis at x = -3, x = 1 and x = 4. Thus (x + 3), (x - 1) and (x - 4) are the 3 factors of the cubic polynomial. Thus the cubic function would be y = k*(x + 3)*(x - 1)*(x - 4). Here we see that the function has a negative y intercept. Let it pass through (0, a) where a is negative.
Thus a = k*3*(-1)*(-4) = k*12, thus k = a/12, since a is negative thus k is negative (k<0).
Thus the function y = k*(x + 3)*(x - 1)*(x - 4), where k<0.
Problem 1.6.27
Whatever be the number of distinct or common roots, for a 5 degree polynomial y would tend to + infinity when x tend to + infinity and y tends to - infinity when x tends to - infinity.
A polynomial function of degree 5, in the first case can have 5 distinct roots. The graph of which will be a curve with y tending to + infinity when x tends to + infinity and y tending to - infinity when x tends to - infinity. Thus the function in this case would turn 4 times during the entire graph.
Next it is also possible that the function cuts the x axis at 3 distinct points and has 2 common roots at a 4th different root. The y value of the function tends to + infinity when x tend to + infinity and y tends to - infinity when x tends to - infinity. The function in this function would again turn 4 times, but in this case, the function would cut the x axis at 3 points and at the 4th point would touch the x axis at 4th point and turn at the same point.
Next another possibility is that we have 2 common roots and another different 2 common roots and another different root, (x - a)^2 * (x - b)^2 * (x - c). The function in this case touches the x axis at x = a and x = b and cuts the x axis at x = c. the function turns 4 times in this case too.
Another possibility will be (x - a)^5. In this case the function cuts the x axis at only one point and the function has the slope of the function 0 at the same point, and the function turns twice around that point.
Next possibility (x - a)^3*(x - b)^2. In this case the function cuts the x axis at the point x = a and touches the x axis at point x = b. thus function thus will have slope value of 0 at points x = a and x = b. The function turns twice around the point x = a and turns at the point x = b.
Thus there are 5 different possibilities for the 5 degree polynomial.
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Another possibility would be 4 common roots with only one distinct.
It is possible that the function approaches -infinity as x -> infinity, and +infinity as x -> -infinity (i.e., coefficient of leading term is negative). However this would not affect the descriptions given here.
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Problem 1.6.29
Given function f(x) = x^3 + 1000x^2 + 1000 and g(x) = x^3 - 1000x^2 - 1000
Naturally for the function to be indistinguishable we will need high values of x and thus corresponding high value of y.
Thus doing so we get the values of x between [- 10^5, 10^5] and the corresponding y value range [ - 10^15 , 10^15 ]
Problem 1.6.32
Given that stopping distance is directly proportional to the square of the velocity.
Let d be the stopping distance and v be the velocity.
When v = 70 mph , d = 177 feet.
Thus d = k*(v^2), where k is the proportionality constant.
Thus 177 feet = k * 70 * 70 square mph
k = 177 / 4900 feet / square mph.
stopping distance corresponding to speed of 35 mph will be
d = 177/4900 feet/square mph * 35 * 35 square mph = 44.25 feet
similarly stopping distance corresponding to speed of 140 mph
d = 177/4900 feet/ square mph * 140 * 140 square mph = 708 feet.
Thus stopping distance corresponding to 140 mph will be 708 feet and to 35mph will be 44.25 feet.
&#Good responses. See my notes and let me know if you have questions. &#