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Question and attempt at question are in the photo below. I have gotten halfway through but I am confused how to show the rest of the question (mainly part c)

Thanks for your help in advance :)

Questions highlighted and attempt below

Kuzja
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erin
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    Note: Use $\phi$ for $\phi$. – marty cohen Oct 29 '14 at 21:56
  • If someone could please help it would be greatly appreciated. if $\phi$'(c)=0 I am unsure how to use this to show u'(c)=L'(c). Is it just as simple as saying $\phi$'(c)=u'(c)-L'(c)=0 So u'(c)=L'(c)? – erin Oct 29 '14 at 22:03

2 Answers2

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You have $2(b)$ right.

$2(c)\;$ Your statement of the theorem is almost right. It needs the restriction that the function must be differentiable at the min/max point $c$. (For an example of why that's needed, consider a V-shaped function: it has a minimum at the bottom of the V but has no derivative at that point.)

Now, $\phi$ is differentiable on $\mathbb{R}$ because both $U$ and $L$ are differentiable on $\mathbb{R}$.

So by your theorem, $\phi'(c) = 0$.

Therefore, \begin{eqnarray*} \phi(x) &=& U(x) - L(x) \\ \phi'(x) &=& U'(x) - L'(x) \\ \phi'(c) &=& U'(c) - L'(c) \\ 0 &=& U'(c) - L'(c) \\ U'(c) &=& L'(c). \\ \end{eqnarray*}

Mick A
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It's not the derivative of $\phi'(c)$ that is zero, but $\phi'(c)$ itself.

marty cohen
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  • sorry thats what I meant. accidentally wrote it out as well as putting in the symbol – erin Oct 29 '14 at 22:07