So I'm doing Mean Value theorem homework which states $$f'(c)=\frac{f(b)-f(a)}{b-a}$$ So I am trying to find $c$ for $f(x)=\sin x$ over the interval $[0,\frac{\pi}{2}]$. So using the Mean Value theorem I got $$f'(c)=\frac{1-0}{\frac{\pi}{2}-0}=\frac{1}{\frac{\pi}{2}}=\frac{2}{\pi}$$ So since $f'(x)=\cos x$ $$\cos(c)=\frac{2}{\pi}$$ but I am having trouble finding the value $c$. Am I right so far, if so how do I find $c$? Thanks in advance.

  • $\begingroup$ What is it that the Mean Value theorem states, exactly? Your expression $f'(c)\frac{f(b)-f(a)}{b-a}$ is not a "statement". $\endgroup$ – Omnomnomnom Jul 14 '14 at 3:33
  • $\begingroup$ It would help if you wrote the problem statement, as it is written $\endgroup$ – Omnomnomnom Jul 14 '14 at 3:34
  • $\begingroup$ I think that is $$ f'(c)=\frac{f(b)-f(a)}{b-a}$$ $\endgroup$ – AsdrubalBeltran Jul 14 '14 at 3:35
  • $\begingroup$ @AsdrubalBeltran thanks, that's what I meant. $\endgroup$ – Kenshin Jul 14 '14 at 3:37
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    $\begingroup$ The answer isn't anything pretty. Assuming they want an answer for the value of $c$, they'll be expecting you to calculate $\cos^{-1}(2/\pi)\approx 0.881$ with a calculator. $\endgroup$ – Omnomnomnom Jul 14 '14 at 3:46

Suppose that the exercise is: Find $c$, for $f(x)=\sin{x}$ with $c\in [0,\frac{π}{2}]$, that satisfied the mean value theorem. Then

With calculator: $c=\cos^{-1}(\frac{2}{\pi})$

Without calculator: how $[0,1]\subseteq[0,\frac{π}{2}]$ you can choose $[0,1]$ then

$$f'(c)=\frac{f(1)-f(0)}{1-0}=\sin(1)$$ then

$$\cos{c}=\sin(1)$$ $$\cos{c}=\cos\left(\frac{\pi}{2}-1\right)$$ $$c=\frac{\pi}{2}-1$$

  • $\begingroup$ Thanks! That was a lot of help. $\endgroup$ – Kenshin Jul 14 '14 at 4:30

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