I tried to solve this problem using the Mean Value Theorem but didn't get very far. This question appeared in my calculus exam and I still can't solve it. Any thoughts?
Thank you.
Suppose $f$differentiable with $f (0) = 0$ and $f (1) = 1$. Show that $f '(X_0)$ = $2X_0$ for some $X_o \in [0,1]$
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calculus
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1**Hint:** Look at $f(x)-x^2$ and use Rolle's theorem. – 2017-01-22
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0Thanks! how did you come up with f(x)-x^2? – 2017-01-22
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0Those equations are usually transferred to a root-finding problem, i.e. $f'(x_0)-2x_0 = 0$. This makes it easier to find solutions. Then i realized that the left side is the derivative of $f(x)-x^2$. – 2017-01-22
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0@LeBtz Thanks a lot that was pretty helpful! – 2017-01-22
1 Answers
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Let $g(x)=f(x)-x^2$ then $g$ is differentiable with $$g(0)=f(0)-0^2=0$$ and $$g(1)=f(1)-1^2=0$$ thus according to the Mean Value Theorem there exists $0<\color{red}{c}<1$ such that $g'(\color{red}{c})=0$ but $g'(x)=f'(x)-2x$ and then $f'(\color{red}{c})-2\color{red}{c}=0$ , finally $f'(\color{red}{c})=2\color{red}{c}$.
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0@LeBtz Yes. Thanks. – 2017-01-22