(1 point) Consider the function 𝑓(𝑥)=7−6𝑥2 on the interval [−6,3]
(A) Find the average or mean slope of the function on this interval, i.e.
𝑓(3)−𝑓(−6)/(3−(−6))=
(B) By the Mean Value Theorem, we know there exists a 𝑐 in the open interval (−6,3) such that 𝑓′(𝑐) is equal to this mean slope. For this problem, there is only one 𝑐 that works. Find it.
𝑐=
To find the average slope of the function on the given interval, we need to calculate the difference in y-values divided by the difference in x-values.
(A) Average slope formula:
Average slope = (f(3) - f(-6)) / (3 - (-6))
To find f(3), substitute x = 3 into the function:
f(3) = 7 - 6(3)^2 = 7 - 6(9) = 7 - 54 = -47
To find f(-6), substitute x = -6 into the function:
f(-6) = 7 - 6(-6)^2 = 7 - 6(36) = 7 - 216 = -209
Plugging the values back into the average slope formula:
Average slope = (-47 - (-209)) / (3 - (-6)) = (-47 + 209) / (3 + 6) = 162 / 9 = 18
Therefore, the average slope of the function on the interval [-6, 3] is 18.
(B) According to the Mean Value Theorem, there exists a c in the open interval (-6, 3) such that f'(c) is equal to the mean slope.
To find this c, we need to find the derivative of the function and set it equal to the mean slope.
The derivative of the function f(x) = 7 - 6x^2 can be found by applying the power rule for differentiation:
f'(x) = -12x
Setting f'(c) equal to the mean slope (18):
-12c = 18
Solving for c:
c = 18 / -12 = -3/2
Therefore, the value of c that satisfies f'(c) = 18 is c = -3/2.