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Prove or contradict:

If $f(x)$ is differentiable in $\mathbb{R}$ and exist $a,b \in \mathbb{R}, a\neq b$ such that $f'(x)=(x-a)(x-b)$ then $f$ has exactly one local minimum and 1 local maximum

I know that $f'$ has only two roots which are $a$ and $b$ as the possible locations for min and max points, but how do I show these are necessarily or are not necessarily min and max points?

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    $\begingroup$ Check the sign of $f'(x)$ when $x<a$, $a<x<b$ and $x>b$. $\endgroup$ – Trevor Norton Apr 24 '18 at 15:57
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    $\begingroup$ Use second derivative test i.e. look at $f''(a)$ and $f''(b)$, whichever is $<0$ is the local maximum and whichever is $>0$ is local minimum. First, without loss of generality, assume $a<b$, then proceed with the test $\endgroup$ – Naweed G. Seldon Apr 24 '18 at 16:00
  • $\begingroup$ mathworld.wolfram.com/SecondDerivativeTest.html $\endgroup$ – user547564 Apr 24 '18 at 16:02
  • $\begingroup$ Alright cool! Got it thanks guys! $\endgroup$ – Jason Apr 24 '18 at 16:05
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\begin{align} f''(x) = \left( x - b \right) + \left( x - a \right) \end{align} \begin{align} f''(a) = \left( a-b \right) \end{align} \begin{align} f''(b) = \left( b -a \right) \end{align} When \begin{align} a \gt b \end{align} Then \begin{align} f''(a) \gt 0 \end{align} And \begin{align} f''(b) \lt 0 \end{align} When \begin{align} b \gt a \end{align} Then \begin{align} f''(a) \lt 0 \end{align} And \begin{align} f''(b) \gt 0 \end{align} As you can see in both cases we have a local Maxima and minima

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  • $\begingroup$ this only proves existence but not uniqueness $\endgroup$ – Surb Apr 24 '18 at 17:59
  • $\begingroup$ When \begin{align} a\gt b \end{align} then \begin{align} f''(a) \gt 0 \end{align} and \begin{align} f''(b) \lt 0 \end{align} so here \begin{align} a \end{align} is the local minima and \begin{align} b \end{align} is local Maxima also when \begin{align} b \gt a \end{align} then \begin{align} f''(b) \gt 0 \end{align} and \begin{align} f''(a) \lt 0 \end{align} so here \begin{align} a \end{align} is local Maxima and \begin{align} b \end{align} is local minima. So in both cases we have a unique Maxima and minima $\endgroup$ – Apurv Apr 25 '18 at 1:01
  • $\begingroup$ First, you may want to know that in Mathjax, $f'(a)$ produces $f'(a)$ and $$f'(a)$$ produces $$f'(a)$$. $\endgroup$ – Surb Apr 25 '18 at 5:36
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    $\begingroup$ So let $ c $ be another local Maxima or Minima then so in that case $ f'(c) = 0 $ and $ (c-b)(c-a) = 0 $ solving this we get $ c=b $ or $ c=a $ which means only Maxima or Minima we can have is $ a $ or $ b $ and when $ a \gt b $ $ a $ is local minima and $ b $ is local Maxima and when $ b \gt a $ $ a $ is Maxima and $ b $ is minima(which I have proven in the answer). So in either case we have a unique Maxima and Minima $\endgroup$ – Apurv Apr 25 '18 at 8:07
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    $\begingroup$ since $f'$ has only two roots (a,b) then those are the only possible maxima and minima points, and since there exists one of each there is exactly one of each. (basically what @Apruv said but less technically) $\endgroup$ – Jason Apr 25 '18 at 8:09
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Let $a<b$ without loss of generality. For $x<a$ we have $f'(x)>0$, for $x=a$ we have $f(a)=0$ and for $a<x<b$ we get $f'(x)<0$. This can be interpreted as $f$ is increasing up to $x=a$ and then decreases while $x$ approaches $b$ from the left. So $f(a)$ is a maximum. Similarly $f'(b)=0$ and $f'(x)>0$ for all $x>b$ which implies that $f$ once it reaches $f(b)$ starts increasing. So $f(b)$ is a minimum. These two points $a,b$ are the only local extrema since $f'(x)$ vanishes exactly there by its very definition.

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  • $\begingroup$ Currently, this is the only correct answer. $\endgroup$ – Surb Apr 24 '18 at 18:01
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The stationary point qualification is done with

$$ f''(x) = (x-a)+(x-b) = 2x-(a+b) $$

hence we have

$$ f''(a) = a-b\\ f''(b) = -(a-b) $$

so one of then is a relative minimum and the other a relative maximum deppending on $a > b$ or $b > a$

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