Find all numbers that satisfy the mean value theorem

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Find all numbers c that satisfy the conclusion of the mean value theorem for the following function and interval:$$f(x)=3x^2+2x+2 \tag{[-1,1]}$$so far I have $$f'(x)=6x+2$$$$6x+2=-1$$$$x=-1/2$$and$$6x+2=1$$$$x=-\frac{1}{6}$$I know both of these are wrong, and the written answer is 0, but I don't see how to get the correct answer.

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2 Answers

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$$ f(1) = 7, f(-1) = 3 $$ Mean value theorem states, for some c in [a, b] with a < b, $$ f'(c) = \frac{f(b)-f(a)}{b-a}, $$ As, you mentioned, $$ f'(x) = 6x + 2 $$ Can you solve now?

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$f(1) = 3\cdot 1^2 + 2\cdot 1 + 2 = 7$.

$f(-1) = 3\cdot (-1)^2 + 2\cdot (-1) + 2 = 3$.

Thus: $f(1) - f(-1) = (1 - (-1))\cdot f'(c)$ gives:

$7 - 3 = 2\cdot f'(c)$.

So: $f'(c) = 2 \to 6c + 2 = 2 \to c = 0$

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