Question

Let f(x)=4x 2

on [−2,1]. Find the value(s) of c that satisfy the conclusion of Mean Value Theorem. 1−(−2)
4−16
​
= 31
−124
​
4 4(1) 2
16
​
4.4) 2
16
​
(a) −2 (b) 2
3
​
​
(c) − 6
5
​
(d) 0 f ′
(x)
e ′
(−2)
​
=−4
=8x
=8(−2)
=−16
​
f ′
(x)
f ′
(1)
​
= 8
8x
​
= 8
−4
​
=8(1)
​
x=− 2
1
​
(C) − 2
1
​

Answer 1

By the mean value theorem, there exists a number c in the interval [-2,1] such that f'(c) = (f(1) - f(-2))/(1 - (-2)).

We have f(x) = 4x^2, so f'(x) = 8x.

Therefore, f'(c) = 8c and

(f(1) - f(-2))/(1 - (-2)) = (4(1)^2 - 4(-2)^2)/(1 - (-2)) = (4-16)/3 = -4/3.

So we need to solve the equation 8c = -4/3 for c in the interval [-2,1].

The only solution in the interval is c = -2/3, so the answer is (e) -2/3.

Explanation: