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If we have a derivable function, $f$, on an interval $[a,b]$, such that :
$f(a)=f(b)=0$ and $f'(a)>0$ and $f'(b)>0$ How de we prove that there exists a $c$ in $]a,b[$ such that : $f(c)>0$ ?

Abbkey
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2 Answers2

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Since the derivative of $f(x)$ is positive at $a$ the function is increasing at that point and $f(a)=0$. Thus there must be some value $c>a$ for which $f(c)>f(a)=0$

aleden
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  • I thought the same but I also thought it was too evident and there must be a "proof" for it. – Abbkey Nov 05 '17 at 20:21
  • It helps to visualize it on a graph. However, if $f'(a)$ were equal to $0$, then we would not know if the function would be increasing or decreasing past that point. – aleden Nov 05 '17 at 20:24
  • Yes I know that. But What if the function is increasing at $a$ but decreasing afterwards ? Would there still be a $c$ ? – Abbkey Nov 05 '17 at 20:29
  • @Abbkey Sure! Think to $\sin x$ on $(0,2\pi)$ – Raffaele Nov 05 '17 at 20:33
  • The function will only start to decrease after achieving a relative maximum. However that only happens after reaching a critical point. Since $f'(a)>0$ it is not a critical point and the function will keep increasing. – aleden Nov 05 '17 at 20:33
  • Alright I got it thank you for helping. – Abbkey Nov 05 '17 at 20:38
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If $f'(a)>0$ then there exist an $h>0$ such that $\dfrac{f(a+h)-f(a)}{h}>0$

that is $f(a+h)>0$. Call $a+h=c$ and you got the proof.

I don't understand very well the second hypothesis, $f'(b)>0$ unless we want to prove that exists at least another point $d\in]a,b[$ such that $f(d)<0$

Indeed if $f'(b)<0$ then there exists an $h<0$ such that $\dfrac{f(b+h)-f(b)}{h}>0$

that is $f(b+h)<0$ and as $h<0$ then $b+h=d\in]a,b[$

I had $\sin x$ on $[0,\pi]$ in mind while writing the proof

Hope this is useful

Raffaele
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  • Absolutely useful. Well yes that's the second part of the question, proving a $d$ u got it right. Thank you for helping. – Abbkey Nov 05 '17 at 20:38