Proving a point exists on a differentiable function

Question

I have a homework question which is:

If $f : [a,b]->R$ is continuous in $[a,b]$ and differentiable at $(a,b)$ and exists a point $c$ in $(a,b)$ such that $(f(c)-f(a))(f(b)-f(c))<0$ then prove that there is a point $d$ in $(a,b)$ such that $f'(d)=0$.

I am having trouble proving this - I am sure I am missing some simple algebra trick to show that $\frac {f(b)-f(a)}{b-a}=0$ or something like that...

Can someone help me please?

Thanks :)

Answer 1

By hypothesis, $f(c)-f(a)$ and $f(b)-f(c)$ must have different signs. So either

1) $f(c)>f(a)$ and $f(b)<f(c)$.

or

2) $f(c)<f(a)$ and $f(b)>f(c)$.

Suppose case 1) holds. Choose any number $r\ne f(c)$ between $f(c)$ and $\max\{f(a),f(b)\}$.

Then on the interval $[a,c]$, we have $f(a)\le r< f(c)$. By the Intermediate Value Theorem, there is a point $d$ in $[a,c]$ with $f(d)=r$. Since $r\ne f(c)$, it follows that $d\in [a,c)$.

On the interval $[c,b]$, we have $f(b)\le r<f(c)$. By the Intermediate Value Theorem, there is a point $e$ in $[c,b]$ with $f(e)=r$. Since $r\ne f(c)$, it follows that $e\in (c,b]$.

Thus we have points $d\ne e$ in $[a,b]$ with $f(d)=f(e)$. The Mean Value Theorem then gives a point $h\in(a,b)$ with $$f'(h)= {f(e)-f(d)\over e-d}= 0.$$

I'll leave case 2) for you.

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In Case 1:

the graph of $f$ intersects the line $y=r$ in each of the intervals $[a,c)$ and $(c,b]$.

Answer 2

Define $\ g(x):= (f(x)−f(a))(f(b)−f(x))$ Clearly g is continuos on $\ [a,b]$ and differentiable (a,b). Note that $\ g(a)=g(b)=0$. By weierstrass g attains its minimum value on the compact interval $\ [a,b]$. Now note that g(c)<0 some c in $\ (a,b) $. By that and the fact that g is zero on the endpoints of the interval it follows that $\ g'(d)=0$ some d in (a,b). But $\ g'(d)=f'(d)(f(a)+f(b)−2f(d))$. But now it follows that either f'(d)=0 in which case we are done, or $\ f(d)=(f(a)+f(b))/2$. But $\ f(d)=f(a)+f(b)/2$ is false since if it was true then $\ g(d)=(f(b)-f(a))^2$ Contradicting that $d \in (a,b) $ is the minimum value attained by $\ g(x).$

Answer 3

$(f(c)-f(a))(f(b)-f(c))<0\implies 2 case f(c)>f(a) and f(b)<f(c) or f(a)>f(c) and f(b)>f(c)$ From both cases,it is easy to use MVT to show the slope of the function f(x) consist of positive and negative and we can concluded that by contiunity,there must exist a number d st $f'(d)=0$