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Let $f$ be a function continuous on $[0, 1]$ and twice differentiable on $(0,1)$.

Suppose that:

$f(0) = f(1) = 0$ and $f(c) >0$ for some $c \in (0,1)$.

Prove that there exists $x_{0}\in(0,1)$ such that $f''(x_{0}) < 0$.

I'm not sure how to approach this question. Do I have to use the “mean value theorem” or is there actually another way to approach this question?

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We can certainly use mean value theorem in it.

Question says $\exists c \in(0,1)$ such that $f(c)>0$. Let $f(c)=\alpha$ and $\alpha>0$

Now from "Lagrange's mean value theorem", $\exists x_{1}\in(0,c)$ such that $$ f'(x_{1}) = \dfrac{f(c)-0}{c-0} = \dfrac{f(c)}{c} $$ Also from "Lagrange's mean value theorem", $\exists x_{2}\in(0,c)$ such that $$ f'(x_{2}) = \dfrac{0-f(c)}{1-c}=\dfrac{-f(c)}{1-c} $$

Here we can claim $x_{1}<x_{2}.$ Also that $f'(x_{1})>0$ and $f'(x_{2})<0$.

We know that derivative of $f'(x)$ is a continuous function and value of $f'(x)$ has decreased while moving from $x_{1}$ to $x_{2}$. So there must be some interval which is the subset of the interval $(x_{1}, x_{2})$ such that f'(x) was decreasing in that interval.

Now, since we are guaranteed of an interval in $(0,1)$ such that $f'(x)$ is decreasing on that interval. At all the points in that interval it is certainly true that $f''(x)<0$.

Hence we are guaranteed of the existence of a point $x\in(0,1)$ such that $f''(x)<0$.

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  • $\begingroup$ If you still need some more help in this question then let me know. $\endgroup$ – Prakhar Gupta Oct 25 '15 at 10:03
  • $\begingroup$ is it possible if i can have a more detailed solution on how to do the question? because i have no clue on how to apply it $\endgroup$ – Jason Tan Oct 25 '15 at 10:07
  • $\begingroup$ i actually have a part (ii) to this question, i will put it in another question. is it okay? $\endgroup$ – Jason Tan Oct 25 '15 at 17:32
  • $\begingroup$ actually i have to ask it here. $\endgroup$ – Jason Tan Oct 25 '15 at 17:33
  • $\begingroup$ Suppose that 􏰄 the definite integral of f(x) dx from 1 to 0 = f(0) = f(1) = 0. Prove that there exists a number x0 ∈ (0,1) such that f′′(x0) = 0. $\endgroup$ – Jason Tan Oct 25 '15 at 17:33
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Let's see what Rolle’s theorem is telling us about $f$. The function has a positive absolute maximum, since it attains a positive value. Such a maximum must be at $t\in(0,1)$, as $f(0)=f(1)=0$. Therefore $f'(t)=0$.

Suppose, by way of contradiction, that $f''(x)\ge0$ for all $x\in(0,1)$. This means that $f'(x)$ is a nondecreasing function on $(0,1)$.

Since $f'(t)=0$, we have $f'(x)\ge0$ for $t<x<1$. Thus $f$ is nondecreasing on $(t,1)$ and therefore…

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