#$&*
course MTH 173
10/23 1:20
Text section 2.1Problem 2.1.1
Given from the table the distance travelled by the car in t = 2 hours, s(2) = 135 km and the distance travelled in t = 5 hours, s(5) = 400 km. Thus the displacement of the car = s(5) - s(2) = 400 km - 135 km = 265 km. The time duration = 5 - 2 = 3 hours.
Thus the average velocity = change in displacement / time interval = 265 km / 3 hours = 88.34 km/hours
Problem 2.1.2
Given from the table, the position of the particle at t = 0 sec is -2 m and at t = 4 is -6 m. the displacement of the body thus will be x(4) - x(0) = -6m - ( -2m) = -4m. The time interval = 4 - 0 = 4 sec.
Thus the average velocity = change in displacement / time interval = - 4m / 4s = - 1m/s.
Problem 2.1.3
Given from the table, the position of the particle at t = 2 is 14 A and at t = 8 is -4 A. The displacement of the body during this interval thus will be x(8) - x(2) = -4 - (14) = -18 A. The time duration is 8 - 2 = 6 sec.
Thus the average velocity = change in displacement / time interval = - 18 A / 6 sec = -3 A/sec
Problem 2.1.6
Given particles distance function s(t) = e^t - 1 where s is in micrometers and t in sec.
Thus distance at t = 2 sec will be s(2) = e^2 - 1 micrometer and distance at t = 4 sec will be s(4) = e^4 -1 micrometer.
Change in distance for the particle = s(4) - s(2) = e^4 - 1 - e^2 + 1 = e^4 - e^2 = 47.2091 micrometer.
Change in time interval = 4 sec - 2 sec = 2 sec
Thus average change in velocity = change in distance / change in time interval = 47.2091 micrometer / 2 seconds = 23.60455 micrometer/sec.
Problem 2.1.7
Distance function s(t) = 4 + 3sin(t) cm
Distance at time t = pi/3, s(pi/3) = 4 + 3sin(pi/3) cm
Distance at time t = 7pi/3, s(7pi/3) = 4 + 3sin(2pi + pi/3) = 4 + 3sin(pi/3) cm
Thus change in distance of the particle - 4 + 3sin(pi/3) - 4 - 3sin(pi/3) = 0 cm.
Change in time interval = 2pi sec
Thus average velocity for that particle in that interval = change in distance / change in time interval = 0/2pi = 0 cm/sec
Problem 2.1.9
Given distance function s = 4 t^3 m
(a) Average velocity between t = 0 sec and t = h sec
(i) Given h = 0.1 sec
Thus distance travelled by the particle in t = 0 sec, s(0) = 0 m
Distance travelled by the particle in t = 0.1 sec = 4 * (0.1)^3 m = 0.004 m
Change in distance for the particle in this interval = 0.004 m - 0 m = 0.004 m
Change in time interval = 0.1 sec - 0 sec = 0.1 sec
Thus average velocity for that interval = change in distance/change in time interval = 0.004 m/0.1 s = 0.04 m/s
(ii) Given h = 0.01 sec
Thus distance travelled by the particle in t = 0 sec, s(0) = 0 m
Distance travelled by the particle in t = 0.01 sec = 4 * (0.01)^3 m = 0.000004 m
Change in distance for the particle in this interval = 0.000004 m - 0 m = 0.000004 m
Change in time interval = 0.01 sec - 0 sec = 0.01 sec
Thus average velocity for that interval = change in distance/change in time interval = 0.000004 m/0.01 s = 0.0004 m/s
(iii) Given h = 0.001 sec
Thus distance travelled by the particle in t = 0 sec, s(0) = 0 m
Distance travelled by the particle in t = 0.001 sec = 4 * (0.001)^3 m = 0.000000004 m
Change in distance for the particle in this interval = 0.000000004 m - 0 m = 0.000000004 m
Change in time interval = 0.001 sec - 0 sec = 0.001 sec
Thus average velocity for that interval = change in distance/change in time interval = 0.000000004 m/0.001 s = 0.000004 m/s
(b) As we see that as h tends to 0, that is as h becomes smaller as smaller from 0.1 sec to 0.01 and finally 0.001 sec, the value of the average rate tends to 0, that is it becomes smaller and smaller from 0.04 m/s to 0.0004 m/s and finally to 0.000004 m/s. The instantaneous velocity change will be the average velocity during the time interval t = 0 and t = h as h tends to 0 sec. Thus the instantaneous velocity of the particle at time t = 0 is 0 m/s
@&
You weren't asked to do this but note also that the expression
( f(h) - f(0) ) / h
represents the average rate of change between x = 0 and x = h.
fIh) = 4 h^3, f(0) = 0 so this expression has value
4 h^3 / h = 4 h^2.
*@
Problem 2.1.11
Given that the a car is driven at a constant speed, which means that the car will travel equal distance in the same interval of time, which means that if the car is at distance x = k km from x = 0 km at t = a from t = 0 sec, at t = 2a sec the car will be at x = 2k km. This characteristics of the car tells that the graph of the distance the car covered as a function of time will be a straight line with a y intercept depending upon the initial distance you consider at the initial start time.
This can also be shown by using integration. The speed function is constant, which is also the average rate at which distance changes with time, thus the distance function would be an anti-derivative of the constant function and thus a linear function.
Problem 2.1.14
To calculate lim h tending to 0 ((3 + h)^3 - 27)/h)
Let f(h) = ((3 + h)^3 - 27)/h)
Thus f(0.1) = 2.791/0.1 = 27.91
f(0.01) = 0.270901/0.01 = 27.0901
f(0.001) = 0.027009/0.001 = 27.009
f(-0.1) = -2.611/-0.1 = 26.11
f(-0.01) = -0.269101/-0.01 = -26.9101
f(-0.001) = -0.026991001/-0.001 = 26.991001
thus as we see that as h tends to 0 from either side the value of the function tends to 27 and thus the limit of the function as h tends to 0 is 27.
Problem 2.1.16
To calculate lim h tending to 0 (7^h -1)/h
Let f(h) = (7^h - 1)/h
Thus f(0.1) = 0.21481/0.1 = 2.1481
f(0.01) = 0.019650/0.01 = 1.9650
f(0.001) = 0.0019478/0.001 = 1.9478
f(-0.1) = -0.176828/-0.1 = 1.76828
f(-0.01) = -0.019271/-0.01 = 1.9271
f(-0.001) = -0.0019442/-0.001 = 1.9442
thus as we see that as h tends to 0 from either side the value of the function tends to a value approximate to 1.94591 which is nothing but ln(7) and thus the limit of the function as h tends to 0 is ln(7).
Problem 2.1.17
To calculate lim h tending to 0 (e^(1+h) - e)/h
Let f(h) = (e^(1+h) - e)/h
Thus f(0.1) = 0.285884/0.1 = 2.85884
f(0.01) = 0.0273191/0.01 = 2.73191
f(0.001) = 0.0027196/0.001 = 2.7196
f(-0.1) = -0.258678/-0.1 = 2.58678
f(-0.01) = -0.027047/-0.01 = 2.70473
f(-0.001) = -0.0027169/-0.001 = 2.7169
thus as we see that as h tends to 0 from either side the value of the function tends to a value approximate to 2.71828 which is nothing but e and thus the limit of the function as h tends to 0 is e.
Problem 2.1.20
Given that the function y = f(x)
As we can see from the graph that the slope of the function at point A is greater than that at the slope of the function at point B which in turn is greater than the slope of the function C. it can also be seen that the slope of the line AB is between the slope of A and B. The slope of the line at point A can be seen to be greater than the slope of the line y = x. Thus the slope of the line y = x is 1 and thus the slope of the function at point A is greater than number 1. Next the slope of the line AB is less than that of the line y = x and thus smaller than the number 1. Since all the slopes discussed above are positive thus they are greater than the number 0. Thus the ascending order of the slope( from the smallest to largest) will be
1. The number 0
2. The slope of the graph at C
3. The slope of the graph at B
4. The slope of the line AB
5. The number 1
6. The slope of the graph at A
Problem 2.1.22
From the table we can see that the position of the car at t = 0 sec is s(0) = 0 ft and at t = 0.2 sec will be s(0.2) = 0.5 ft.
Thus the change in the distance = s(0.2) - s(0) = 0.5 ft
Change in time interval = 0.2 - 0 = 0.2 sec
Thus the average velocity over that interval = change in distance / change in time interval = 0.5 ft / 0.2 sec = 2.5 ft/sec.
In order to estimate the velocity at t = 0.2 sec we find average velocity and then see what the value tends for smaller interval.
For interval of time 0.2 sec to 1.0 sec
Change in distance = s(1.0) - s(0.2) = 9.6 ft - 0.5 ft = 9.1 ft
Change in time duration = 1.0 - 0.2 = 0.8 sec
Average velocity = change in distance / change in time duration = 9.1 ft / 0.8 sec = 11.375 ft/sec
For interval of time 0.2 sec to 0.8 sec
Change in distance = s(0.8) - s(0.2) = 6.5 ft - 0.5 ft = 6 ft
Change in time duration = 0.8 - 0.2 = 0.6 sec
Average velocity = change in distance / change in time duration = 6 ft / 0.6 sec = 10 ft/sec
For interval of time 0.2 sec to 0.6 sec
Change in distance = s(0.6) - s(0.2) = 3.8 ft - 0.5 ft = 3.3 ft
Change in time duration = 0.6 - 0.2 = 0.4 sec
Average velocity = change in distance / change in time duration = 3.3 ft / 0.4 sec = 8.25 ft/sec
For interval of time 0.2 sec to 0.4 sec
Change in distance = s(0.4) - s(0.2) = 1.8 ft - 0.5 ft = 1.3 ft
Change in time duration = 0.4 - 0.2 = 0.2 sec
Average velocity = change in distance / change in time duration = 1.3 ft / 0.2 sec = 6.5 ft/sec
As we see that as the interval gets smaller and smaller the value of the average velocity decreases from 11.375 to 6.5 ft/sec. Since from the data available to us the smallest interval is of 0.2. Thus if we consider from the right hand side the average velocity for the interval t = 0.2 sec to t = 0.4 sec is 6.5 ft/sec.
If we consider from the left hand side the average velocity for the interval t = 0 sec and t = 0.2 sec is 2.5 ft/sec.
Thus the estimation of the velocity at t = 0.2 sec will be the average of the average velocity obtained from the right hand side of 0.2 and left hand side.
Thus the estimation of average velocity at t = 0.2 sec = (2.5 + 6.5)/2 = 4.5 ft/sec
Text section 2.2
Problem 2.2.2
Given function f(x) = x^3
We need to figure the instantaneous rate of change of the function at x =1. Let h be a small interval around x = 1. The instantaneous rate of change of the function will be f( 1 + h) - f(1) / h provided h is tending to 0. Thus we do by using small values of h.
Let h = 0.1, thus the instantaneous rate of change = f(1.1) - f(1) / 0.1 = 0.331/0.1 = 3.31
Let h = 0.01, instantaneous rate of change = 0.030301/0.01 = 3.0301
Let h = 0.001, instantaneous rate of change = 0.003003001/0.001 = 3.003001
Let h = -0.1, instantaneous rate of change = -0.271/-0.1 = 2.71
Let h = -0.01, instantaneous rate of change = -0.029701/-0.01 = 2.9701
Let h = -0.001, instantaneous rate of change = -0.002997001/-0.001 = 2.997001
As we see that as the x value of the function tends to 1 from both the sides the value of the function tends to the value of 3. Thus the limit of the function as x tends to 1 is 3
Problem 2.2.3
(a) revenue in terms of quantity sold R(q) = 100q - 10(q^2)
First time interval of q between 1 and 2.
The revenue at q = 1 kg, R(1) = 100 - 10 = $90
Revenue at q = 2 kg, R(2) = 200 - 40 = $160
Change in the revenue = $160 - $90 = $70
Change in q = 2 - 1 = 1 kg
Average rate of change of R with respect to q over 1 < q < 2 = $70/1kg = $70 / kg
For time interval of q between 2 and 3
The revenue at q = 2 kg, R(2) = $160
The revenue at q = 3 kg, R(3) = 300 - 90 = $210
Change in the revenue = $210 - $160 = $50
Change in q = 3 - 2 = 1 kg
Average rate of change of R with respect to q over 2 < q < 3 = $50 / kg
(b) let h be small increment in x = 2 to find the instantaneous rate of change of revenue. By choosing small value of h. we estimate the instantaneous rate of change of revenue with respect to change in quantity at q = 2 kg
Instantaneous rate of change of revenue at q = 2 kg = (R( 2 + h) - R(2))/h , h tending to 0.
Let f(h) = (R(2 + h) - R(2))/h
Let h = 0.1, f(0.1) = (R(2.1) - R(2))/0.1 = 5.9/0.1 = $59/kg
For h = 0.01, f(0.01) = 0.599/0.01 = $59.9/kg
For h = 0.001,f(0.001) = 0.05999/0.001 = $59.99/kg
For h = -0.1, f(-0.1) = -6.1/-0.1= $61/kg
For h = -0.01, f(-0.01) = -0.601/-0.01 = $60.1/kg
For h = -0.001, f(-0.001) = -0.06001/-0.001 = $60.01/kg
Thus as we see that as we tend the quantity sold value to 2 by using small values of h on both sides the value of the function tends t0 $60/kg, which must therefore be the limit of the revenue function at quantity 2 kg.
Problem 2.2.4
(a) given f(x) = e^x
For x = 1, f(1) = e^1 = 2.718
For x = 1.5, f(1.5) = e^1.5 = 4.482
For x = 2, f(2) = e^2 = 7.389
For x = 2.5, f(2.5) = e^2.5 = 12.182
For x = 3, f(3) = e^3 = 20.086
(b) the average rate of change of the function between x = 1 and x = 3 will be (f(3) - f(1))/(3-1)
= (20.086 - 2.718) / 2 = 8.684
(c) to calculate the approximate instantaneous rate of change at x = 2 we use the smallest interval available on either side of x = 2 that is of 0.5.
Thus for the interval of x = 1.5 to x = 2, the average rate of change of function
= ( 7.389 - 4.482 ) / 0.5 = 5.814
Thus the average rate of change of function for the interval on the left hand side of x = 2 will be 5.814
For the interval of x = 2 to x = 2.5, the average rate of change of function
= ( 12.182 - 7.389 ) / .5 = 9.586
Thus the accurate estimation of the instantaneous rate at x = 2 will be the average of the average rate of change on either side x = 2.
Thus the instantaneous rate at x = 2 will be (9.586 + 5.814)/2 = 7.7
Problem 2.2.5
(a) Given function f(x) = log(x)
At x = 1, f(1) = log(1) = 0
At x = 1.5, f(1.5) = log(1.5) = 0.1761 approx
At x = 2, f(2) = log(2) = 0.3010 approx
At x = 2.5, f(2.5) = log(2.5) =0.3979 approx
At x = 3, f(3) = log(3) = 0.4771 approx
(b) the value of the function at x = 1 is 0 and at x = 3 is 0.4771
Change in value of the function = 0.4771 - 0 = 0.4771
Change in x = 3 - 1 = 2
Thus the average rate of change of f(x) between x = 1 and x = 3 thus will be = ( f(3) - f(1) )/2 = 0.4771/2 = 0.23855
(c) To calculate the approximate instantaneous rate of change at x = 2 we use the smallest interval available on either side of x = 2 that is of 0.5.
Thus for the interval of x = 1.5 to x = 2, the average rate of change of function
= ( 0.3010 - 0.1761 ) / 0.5 = 0.2498
Thus the average rate of change of function for the interval on the left hand side of x = 2 will be 5.814
For the interval of x = 2 to x = 2.5, the average rate of change of function
= (0.3979 - 0.3010 ) / .5 = 0.1938
Thus the accurate estimation of the instantaneous rate at x = 2 will be the average of the average rate of change on either side x = 2.
Thus the instantaneous rate at x = 2 will be (0.2498 + 0.1938)/2 = 0.2218
Problem 2.2.6
Given f(x) = x^3 + 4x
To calculate the approximate value of f’(3) we use value in the nearby space of 0.001.
Thus within the interval of 2.999 to 3.001
The value of the function at x = 3, f(3) = 39
The value of the function at x = 2.999, f(2.999) = 38.969
The value of the function at x = 3.001, f(3.001) = 39.031
For the interval of x = 2.999 to x = 3, the average rate of change of function
= ( 39 - 38.969 ) / 0.001 = 31
For the interval of x = 3 to x = 3.001, the average rate of change of function
= (39.031 - 39) / 0.001 = 31
As we can see that that the average rate of function change from both sides of x = 2 is 31, thus the approximate value of the rate of change of the function at x = 3 will be 31
Problem 2.2.7
Given f(x) = sin(x). From the graph it can be clearly seen that the value of the function at x = 3pi is 0 and the function is decreasing at that point. The derivative of the function at a point is nothing but the slope of the function at the same point. Thus to decide whether the derivative of the function is positive or negative we need to see whether the slope of the function is negative or positive. Since the function is decreasing at the point x = 3pi, thus the slope of the function is negative and so the derivative of the function f(x) at x = 3pi is negative.
Problem 2.2.9
Given f(x) = 3^x
To estimate the value of f’(2) we use the values near x = 2 spaced by 0.001. Accuracy up to 3 significant figures is used, the closer spaced digits are used the more accurate the result it.
Thus we use an interval of 1.999 to 2.001.
The value of the function at x = 2, f(2) = 3^2 = 9
The value of the function at x = 1.999, f(1.999) = 3^1.999 = 8.99011
The value of the function at x = 2.001, f(2.001) = 3^2.001 = 9.00989
For the interval of x = 1.999 to x = 2, the average rate of function change will be
= (9 - 8.99011)/0.001 = 9.89
For the interval of x = 2 to x = 2.001, the average rate of function change will be
= (9.00989 - 9)/0.001 = 9.89
Thus as we see that the value of the average rate of change of the function on either side of x = 2 is 9.89, thus the approximate value of the instantaneous rate of change of the function at x = 2 or the derivative of the function at x = 2 will be 9.89.
Problem 2.2.10
(a) Average rate of change between x = 1 and x = 3 is the slope of the line joining the point corresponding on the graph to x = 1 and x = 3. Similarly average rate of change between x = 3 and x = 5 is the slope of the line joining the point corresponding on the graph to x = 3 and x = 5. It is clearly visible from the graph that the slope of the line joining the points corresponding to x = 1 and x = 3 is greater than the slope of the line joining the points corresponding to x = 3 and x = 5 on the graph and thus the average rate of change of f(x) between x = 1 and x = 3 is greater than that between x = 3 and x = 5.
(b) the graph of the function is always increasing. Since the function is always increasing, the value of f(5) will be greater than f(2) as 5 is greater than 2.
(c) f’(1) is the slope of the graph at x = 1 and f’(4) is the slope of the graph at x = 4. Since the function is increasing at a decreasing rate thus the slope of the function is decreasing as x increases thus f’(1) will be greater than f’(4) as 4 is greater than 1.
Problem 2.2.13
Given f(100) = 35
Slope of the function at x = 100, f’(100) = 3.
The value of the function that is f(102) can be estimated by using the slope of the line at x = 100.
Slope = rise / run.
Run for the slope triangle assumed = 102 - 100 = 2
Thus the rise for that triangle = 2 * slope of the function at x = 100 = 2 * f’(100) = 2 * 3 = 6
The value of the function at x = 102 = f(102) = rise + f(100) = 6 + 35 = 41.
Thus the estimated result of the function at x = 102, f(102) = 41
Problem 2.2.16
As we can see from the graph of the function, the function is ever increasing at a decreasing rate. This means that the slope of the function is decreasing as x increases but the slope remains positive throughout.
Thus the slope of the function at x = 2 will be greater than the slope of the function at x = 3.
Thus f’(2) > f’(3)
f(3) - f(2) can also be written as (f(3) - f(2))/(3 - 2), which is nothing but the average rate of change of the function between x = 2 and x = 3, which is the slope of the line joining x = 2 and x = 3. Since it is the average slope thus its value will be between the slope of the function at x = 2 and the slope of the function at x = 3. Since the slope is decreasing thus we have
f’(2) > f(3) - f(2) > f’(3).
Since all the slopes calculated above are positive, they are greater than the number 0 thus the correct ascending order will be
0 < f’(3) < f(3) - f(2) < f’(2).
Problem 2.2.17
Given f(4) = 25 and f’(4) = 1.5. The x coordinate of the point A is 4 and the value of the function corresponding to x = 4 will be 25. Thus the coordinate of A is ( 4 , 25 )
To calculate the estimate coordinate of B we use the slope triangle joining point A and B.
Slope = rise/run. The slope of the triangle is given f’(4) = 1.5. the run for the triangle will be = 4.2 - 4 = .2
Rise for the triangle = slope * run = 1.5 * .2 = 0.3
Thus the value of the function corresponding to x = 4.2, f(4.2) = f(4) + rise = 25 + .3 = 25.3
Thus the estimated coordinate of B (4.2,25.3)
To calculate the estimate coordinate of C we use the slope triangle joining point C and A.
Slope = rise/run. The slope of the triangle is given f’(4) = 1.5. the run for the triangle will be = 4 - 3.9 = .1
Rise for the triangle = slope * run = 1.5 * .1 = 0.15
Thus the value of the function corresponding to x = 3.9, f(3.9) = f(4) - rise = 25 - .15 = 24.85
Thus the estimated coordinate of C (3.9,24.85)
Problem 2.2.20
(a) f(x) will be the value of the function at any x on the graph
(b) f(x + h) will be the value of the function at a point on the right of x close enough to x on the graph
(c) f(x + h) - f(x) will be the difference of the value of the function at the point x + h and x. If we construct a triangle below the graph of the function by joining (x,f(x)) , ( x + h, f(x + h)) and (x, f(x + h)). The height of this triangle will be f(x + h) - f(x)
(d) h will be the run for that triangle, that is increase in the x coordinate.
(e) thus (f(x + h) - f(x))/ h is the slope of that triangle and in turn the average slope of the function between x and x + h. if h tends to 0 the average slope becomes instantaneous slope of the line on the graph at point x.
Problem 2.2.23
Since g is an odd function, the derivative g’(x) will be an even function. for an even function g’(x) = g’(-x)
Thus is g’(4) = 5, g’(-4) will also be 5.
Problem 2.2.27
Derivative of the function f(x) = x^x at x = 2. To find that we use the values of the function in the vicinity of x = 2. We find values of the function for x values within 0.001 spacing of x = 2 and use the average rate to estimate the actual value.
Thus we see the function within the interval of x = 1.999 to x = 2.001
The value of the function at x = 2, f(2) = 2^2 = 4
The value of the function at x = 1.999, f(1.999) = 1.999^1.999 = 3.99323
The value of the function at x = 2.001, f(2.001) = 2.001^2.001 = 4.00677
Thus for the interval of x = 1.999 to x = 2, the average rate of change of function
= (4 - 3.99323)/0.001 = 6.77
For the interval of x = 2 to x = 2.001, the average rate of change of function
= (4.00677 - 4)/0.001 = 6.77
As we see that the average rate of function change from both the left and the right side of x = 2 os 6.77, thus the approximate value of the derivative of the function x^x at x = 2 is 6.77.
Problem 2.2.29
To calculate the instantaneous rate of change of function f(x) = ln(cos(x)) at x =1 and at x = pi/4
To calculate the approximate instantaneous rate of change of function at x = 1 we use the values of the function in the vicinity of x = 1 by space of 0.001, approximating the result up to 3 significant figures.
Thus we use the x interval of x = 0.999 to x = 1.001
The value of the function at x = 1, f(1) = log(cos(1)) = -.61562
The value of the function at x = 0.999, f(.999) = log(cos(.999)) = -.61407
The value of the function at x = 1.001, f(1.001) = log(cos(1.001)) = -.61718
For the interval from x = 0.999 to x = 1, the average rate of change of function will be
= (-.61562 + .61407) / 0.001 = -1.55
For the interval from x = 1 to x = 1.001, the average rate of change of function will be
= (-.61718 + -.61562) / 0.001 = -1.56
As we see that the average rate of change of the function for the interval on the right hand side of x = 1 is -1.56 and that on the left hand side is -1.55, thus the approximate instantaneous rate of function change at x = 1 will be the average of the average rate of function on both the side which is
= (-1.56 - 1.55 ) / 2 = -1.555
To calculate the approximate instantaneous rate of change of function at x = pi/4 = 0.785 ( up to 3 significant figures) we use the values of the function in the vicinity of x = 0.785 by space of 0.001, approximating the result up to 3 significant figures.
Thus we use the x interval of x = 0.784 to x = 0.786
The value of the function at x = 0.785, f(0.785) = log(cos(0.785)) = -.34617
The value of the function at x = 0.784, f(.784) = log(cos(.784)) = -.34517
The value of the function at x = 0.786, f(.786) = log(cos(.786)) = -.34717
For the interval from x = 0.784 to x = .785, the average rate of change of function will be
= (-.34617 + .34517 ) / 0.001 = -1
For the interval from x = .785 to x = .786, the average rate of change of function will be
= (-.34717 + -.34617) / 0.001 = -1.
As we see that the average rate of change of the function on either side of x = pi/4 is the same and equal to -1, thus we can approximate the value of the instantaneous rate of change of the function at x = pi/4 as -1.
Problem 2.2.32
(a)
Given f(x) = (1/2)*(x^2)
And g(x) = f(x) + 3
For both the functions f(x) and g(x) it can be said about the tangents for both the function that the slope of the tangent at x = 0 is same for both g(x) and f(x)
Similarly for x = 2, the slope of tangent at the same x coordinate will still be the same for g(x) and f(x)
Actually for any x = x0 the slope of the tangent remains the same for f(x) and g(x)
(b)
The function g(x) differs from f(x) only by the constant. g(x) function will be like f(x) function on graph shifted by 3 units in the positive y direction. Thus there is no difference in the shape of f(x) and g(x) function and thus no change in slope of both the function at any x. Thus by adding a constant we just shift the function along the y axis without altering the shape that is the slope of the function. Thus the slope of the function is same for the functions f(x) and g(x) = f(x) + C, where C is any constant.
Problem 2.2.35
To calculate lim h tending to 0 (( -3 + h )^2 - 9)/h
Using algebra we reduce the given function into determinant form.
(( -3 + h)^2 - 9)/h = ( h^2 + 9 -6h - 9)/h = (h^2 - 6h)/h = h - 6
Thus lim h tending to 0 (( -3 + h )^2 - 9)/h = lim h tending to 0 h - 6 = -6
Problem 2.2.38
To calculate lim h tending to 0 ((1/(1 + h)^2) - 1)/h
Using algebra we reduce the given function into determinant form.
((1/(1 + h)^2) - 1)/h = ( 1 - (1 + h)^2)/(h*(1 + h)^2) = ( 1 - 1 - h^2 - 2h)/(h*(1 + h)^2) =-( h + 2 ) / (1+h)^2
Thus lim h tending to 0 ((1/(1 + h)^2) - 1)/h = lim h tending to 0 -( h + 2 ) / (1+h)^2 = -2
Problem 2.2.41
Given f(x) = 5x^2
The derivative function f’(x) = lim h tending to 0 (f( x + h) - f(x))/h
f’(x) = lim h tending to 0 ( 5*(x + h)^2 - 5x^2)/h
using algebra we solve the function to find the derivative
f’(x) = lim h tending to 0 5*( 2xh + h^2 )/h
= lim h tending to 0 10x + 5h = 10x
Thus the derivative function will be f’(x) = 10x
The derivative of the function at x = 10 will be f’(10)
f’(10) = 10*10 = 100
Thus the derivative of the function at x = 10 is 100.
"
@&
Very good.
Check my one note.
*@