How to find the value of $c$ using the mean value theorem?

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So I'm doing Mean Value theorem homework which states $$f'(c)=\frac{f(b)-f(a)}{b-a}$$

I have $f(x)=e^{\frac{-x}{2}}$ over the interal [0,12]. Using the mean value theorem I have$$f'(c)=\frac{e^{-6}-e^0}{12}=-.0831267707$$ So for $f'(x)= f'(x)=\frac{-e^{\frac{-x}{2}}}{2}$. So I need to find the exact value for which $f'(x)=\frac{-e^{\frac{-x}{2}}}{2}-.0831267707$ Which I am having problems with. Thanks for all the help! Here's a pic of the actual question.

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1 Answer

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Like said, your derivative is wrong,

$$\frac{d}{dx}f(x) = -\frac{1}{2e^{x/2}}$$

Going back to the MVT,

$$f'(c)=\frac{f(b)-f(a)}{b-a}$$ $$-\frac{1}{2e^{c/2}} = \frac{e^{-6} - 1}{12}$$ $$-\frac{6}{e^{c/2}} = {e^{-6} - 1}$$ $$-6 = e^{c/2}*(e^{-6}-1)$$ $$e^{c/2} = -\frac{6}{e^6-1}$$ $$\ln(e^{c/2}) = \ln(\frac{-6}{e^{-6}-1})$$ $$c/2 = \ln(\frac{-6}{e^{-6}-1})$$ $$c = 2\ln(\frac{-6}{e^{-6}-1})$$

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