10. consider the graph of y = f(x) provided in figure 1.3.10.\na. on the graph of y = f(x), sketch and label…

10. consider the graph of y = f(x) provided in figure 1.3.10.\na. on the graph of y = f(x), sketch and label the following quantities:\n - the secant line to y = f(x) on the interval -3, -1 and the secant line to y = f(x) on the interval 0, 2.\n - the tangent line to y = f(x) at x = -3 and the tangent line to y = f(x) at x = 0.\nb. what is the approximate value of the average rate of change of f on -3, -1? on 0, 2? how are these values related to your work in (a)?\nc. what is the approximate value of the instantaneous rate of change of f at x = -3? at x = 0? how are these values related to your work in (a)?

10. consider the graph of y = f(x) provided in figure 1.3.10.\na. on the graph of y = f(x), sketch and label the following quantities:\n - the secant line to y = f(x) on the interval -3, -1 and the secant line to y = f(x) on the interval 0, 2.\n - the tangent line to y = f(x) at x = -3 and the tangent line to y = f(x) at x = 0.\nb. what is the approximate value of the average rate of change of f on -3, -1? on 0, 2? how are these values related to your work in (a)?\nc. what is the approximate value of the instantaneous rate of change of f at x = -3? at x = 0? how are these values related to your work in (a)?

Answer

Explanation:

Step1: Recall secant - line concept

The secant line between two points ((x_1,y_1)) and ((x_2,y_2)) on (y = f(x)) is a straight - line connecting these two points. For the interval ([-3,-1]), find the points ((-3,f(-3))) and ((-1,f(-1))) on the graph of (y = f(x)) and draw a straight line between them. Similarly, for the interval ([0,2]), find the points ((0,f(0))) and ((2,f(2))) and draw a straight line between them.

Step2: Recall tangent - line concept

The tangent line to (y = f(x)) at (x=a) is the line that touches the curve (y = f(x)) at the point ((a,f(a))) and has the slope equal to (f^{\prime}(a)). For (x=-3), draw a line that just touches the curve at the point ((-3,f(-3))). For (x = 0), draw a line that just touches the curve at the point ((0,f(0))).

Step3: Calculate average rate of change

The average rate of change of (y = f(x)) on the interval ([x_1,x_2]) is given by (\frac{f(x_2)-f(x_1)}{x_2 - x_1}). For the interval ([-3,-1]), the average rate of change is (\frac{f(-1)-f(-3)}{-1-(-3)}=\frac{f(-1)-f(-3)}{2}). For the interval ([0,2]), the average rate of change is (\frac{f(2)-f(0)}{2 - 0}=\frac{f(2)-f(0)}{2}). The slope of the secant line on an interval ([x_1,x_2]) is equal to the average rate of change of the function on that interval.

Step4: Estimate instantaneous rate of change

The instantaneous rate of change of (y = f(x)) at (x=a) is (f^{\prime}(a)), which is the slope of the tangent line at (x = a). To estimate (f^{\prime}(-3)), we look at the slope of the tangent line at (x=-3). To estimate (f^{\prime}(0)), we look at the slope of the tangent line at (x = 0).

Answer:

a. Sketch the secant and tangent lines as described above. b. The average rate of change on ([-3,-1]) is (\frac{f(-1)-f(-3)}{2}), and on ([0,2]) is (\frac{f(2)-f(0)}{2}). These values are equal to the slopes of the respective secant lines. c. The instantaneous rate of change at (x=-3) is (f^{\prime}(-3)) (slope of the tangent line at (x=-3)) and at (x = 0) is (f^{\prime}(0)) (slope of the tangent line at (x = 0)). These values are the slopes of the respective tangent lines.