Question

Conisider the function f(x)=4−5x²on the interval (−4,5] (A) Find the average or mean slope of the function on this interval, i.e.f(5)−f(−4)/5−(−4)​ (B) By the Mean Value Theorem, wo know there exists a c in the open interval (−4,5) such that f′(c) is equal to this mean slope. For ths problem there it only one c that works Find it.

Answer 1

The mean slope of the function f(x) = 4 - 5x² on the interval (-4, 5] is -37/9. By the Mean Value Theorem, we find that the value of c in the interval (-4,5) where f'(c) equals the mean slope is 37/90 or approximately 0.4111.

Step-by-step explanation:

Calculating the Mean Slope and Using the Mean Value Theorem

To find the average or mean slope of the function f(x) = 4 - 5x² on the interval (-4, 5], we need to calculate the change in the function value over the interval:

f(5) - f(-4) / (5 - (-4))

We plug in the values to get:
(4 - 5(5)²) - (4 - 5(-4)²) / 5 - (-4) = (-121 - (-84)) / 9 = (-121 + 84) / 9 = -37 / 9.

The mean slope is -37/9.

According to the Mean Value Theorem, there exists a c in the interval (-4,5) where the function's derivative, f'(c), equals the mean slope.

First, we find the derivative:

f'(x) = -10x

Then we solve for c when f'(c) = -37/9:
-10c = -37/9
c = (37/9) / 10 = 37/90

So, the value of c that satisfies the theorem is approximately 0.4111.