The value of x in the interval -3 ≤ x ≤ 0, where the instantaneous rate of change of h equals the average rate of change of f, is approximately x ≈ -2.15.
Find the value of x in the interval -3 ≤ x ≤ 0 where the instantaneous rate of change of h equals the average rate of change of f over the interval -3 ≤ x ≤ 0:
1. Find the average rate of change of f over the interval -3 ≤ x ≤ 0.
The average rate of change of f over the interval is given by the slope of the secant line that intersects the graph of f at the endpoints of the interval. From the table, we can see that the y-coordinates of the endpoints are f(-3) = -5 and f(0) = 7. Therefore, the slope of the secant line is:
m = (f(0) - f(-3)) / (0 - (-3)) = (7 - (-5)) / 3 = 4
2. Find the instantaneous rate of change of h.
The instantaneous rate of change of h at a point x is given by the value of its derivative h'(x). From the table, we can see that h'(x) = 4e^cos x. Therefore, we need to find the value of x in the interval -3 ≤ x ≤ 0 for which h'(x) = 4.
Unfortunately, there is no closed-form solution for the equation h'(x) = 4. This means that we cannot isolate x algebraically. However, we can use numerical methods to approximate the solution. One way to do this is to use a graphing calculator or computer software to plot the graph of h'(x) and look for the intersection point with the line y = 4.
3. Approximate the solution using a graphing calculator or computer software.
Do this on a graphing calculator:
Enter the function h'(x) = 4e^cos x into the calculator.
Set the window settings to include the interval -3 ≤ x ≤ 0.
Graph the function h'(x) and the line y = 4.
Use the калькулятор intersection feature to find the approximate x-coordinate of the intersection point.
You should find that the intersection point occurs at approximately x = −2.15.
Therefore, the instantaneous rate of change of h equals the average rate of change of f over the interval -3 ≤ x ≤ 0 at x ≈ −2.15.
Complete question:
Let h be the function with derivative given by h'(x) = 4e^cos(x). At what value of x in the interval -3 ≤ x ≤ 0 does the instantaneous rate of change of h equal the average rate of change of f over the interval 3 ≤ x ≤ 0.