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The problem states: Suppose $f'(b) = M$ and $M <0$. Find $\delta>0$ so that if $x\in (b-\delta, b)$, then $f(x) > f(b).$

This intuitively makes sense, but I am not exactly sure how to find $\delta$. I greatly appreciate any help I can receive.

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    I changed $\epsilon-\delta$ to $\epsilon$-$\delta$. When the hyphen is _inside_ the $\TeX$ code, then it gets rendered as a minus sign rather as a hyphen.2012-11-12

2 Answers 2

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Remember that the definition of derivative will imply that $$ \lim_{x\to b^-}\frac{f(b)-f(x)}{b-x}=M. $$ But, $M<0$ and $b-x>0$.

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    If it were not for the fact that I had to find delta, that would definitely work.2012-11-12
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    You find delta based on the definition of the limit. Rattle off the definition: For every $\epsilon>0$ there is a $\delta>0$... and there you go, you "found" $\delta$.2012-11-12
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    @user42864 To clarify nayrb's comment. The limit in my answer carries with it an $\epsilon-delta$ definition. Since $M$ is negative, we can find a $\delta$ so that $\frac{f(b)-f(x)}{b-x}$ is negative. The details are for you to fill in.2012-11-12
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In general you won't be able to 'find $\delta$' explicitly with only information about $f^\prime$, you can only guarantee its existence based on the the limit definition of the derivative. This is because the value of $\delta$ will depend on the curvature of the function at $b$, i.e. the second derivative (which, unless explicitly stated, doesn't necessarily exist).