Find c in the interval [-2,1] which satisfies the mean value theorem for f(x)=x^3-2x^2+x+1

Answer 1

To apply the mean value theorem, we need to find a value c in the interval [-2, 1] such that f'(c) = (f(1) - f(-2))/(1 - (-2)).

First, let's find f'(x):
f'(x) = d/dx (x^3 - 2x^2 + x + 1)
= 3x^2 - 4x + 1

Next, let's find f(1) and f(-2):
f(1) = (1)^3 - 2(1)^2 + (1) + 1
= 1 - 2 + 1 + 1
= 1

f(-2) = (-2)^3 - 2(-2)^2 + (-2) + 1
= -8 - 8 - 2 + 1
= -17

Now, let's evaluate the derivative f'(x) at different values of x within the interval [-2, 1]:

f'(-2) = 3(-2)^2 - 4(-2) + 1
= 12 + 8 + 1
= 21

f'(1) = 3(1)^2 - 4(1) + 1
= 3 - 4 + 1
= 0

The mean value theorem states that there exists a c in the interval (-2, 1) such that f'(c) = (f(1) - f(-2))/(1 - (-2)) = (1 - (-17))/(1 - (-2)) = 18/3 = 6.

Therefore, c = 6 satisfies the mean value theorem for the function f(x) = x^3 - 2x^2 + x + 1 in the interval [-2, 1].

Answer 2

but c=6 is not in the interval [-2,1]

you want c such that f'(c) = 6
That is,
3c^2 - 4c + 1 = 6
c = (2±√19)/3
so we want c = (2-√19)/3 ≈ -0.7863
which is in the interval [-2,1]

Answer 3

Apologies for the mistake. You are correct. The correct value of c that satisfies f'(c) = 6 in the interval [-2, 1] is c = (2-√19)/3 ≈ -0.7863. Thank you for catching that mistake.