1
$\begingroup$

$ Let \ f: [-1,1] \rightarrow \mathbb{R} $ be a function, which is continuous on [-1,1] and twice differentiable on (-1,1) with: $$ f(-1)=f(1)= 0 \ and \ f(0)=-3 $$ Show that there exists $ c \ within \ (-1,1) $ s.t. $ f''(c) > 3 $

My intuition was to use Rolle's theorem to show that there must be a K within (-1,1) s.t. f'(k) = 0, then use prove by contradiction and use mean value theorem to show that there is a f''(c) which cant be $ \leq 3$ however, i am having serious trouble with this. I think i am overseeing something simple, I also figure i need to use the fact that $ f(0)=-3 $ somehow but can't work out where. Some hints would be greatly appreciated!

$\endgroup$
1
$\begingroup$

Hint: Instead of using the mean value theorem (in the form of Rolle's theorem) for $f$ on all of $[-1,1]$, use it separately on each of the intervals $[-1,0]$ and $[0,1]$.

$\endgroup$
2
  • $\begingroup$ Would you then split the proof up into 2 cases, one where K (from Rolle's) is in [-1,0] and the other case when it is in [0,1] ? $\endgroup$ – Matt_G Oct 8 '18 at 5:38
  • $\begingroup$ You might be able to do that, but that makes it more complicated than necessary. You don't need to use Rolle's theorem at all. $\endgroup$ – Eric Wofsey Oct 8 '18 at 5:44
1
$\begingroup$

$-3=\frac{f(-1)-f(0)}{-1\ -0}=f'(a)$ for some $a\in(-1,0)$ and $3=\frac{f(1)-f(0)}{1-0}=f'(b)$ for some $b\in(0,1)$ . Now $\frac{f'(b)-f'(a)}{b-a}=f''(c)$ for some $c\in(a,b)$, also $0<b-a<2$.Therefore $f''(c)>3$

$\endgroup$
1
$\begingroup$

One can even show that $f''(x) \ge 6$ for some $c \in (-1, 1)$ under the given conditions. (The function $f(x) =3(x^2-1)$ shows that this is the best possible bound.)

Assume that $f''(x) < 6$ for all $x \in (-1, 1)$.

Let $x_0 \in (-1, 1)$ be a point where $f$ assumes it's minimum, then $f(x_0) \le -3$ and $f'(x_0) = 0$. Because of the symmetry of the problem we can assume that $x_0 \ge 0$.

Taylor's formula applied to the interval $[x_0, 1]$ gives $$ f(1) = f(x_0) + f'(x_0)(1-x_0) + \frac{f''(\xi)}{2}(1-x_0)^2 $$ for some $\xi \in (x_0, 1)$, and therefore $$ 0 = f(1) < -3 + \frac 62 (1-x_0)^2 \le -3 + 3 = 0 $$ which is a contradiction.


An alternative (mimicking the proof of Taylor's theorem) is to consider the function $$ g(x) = f(x) + (x^2 -1) \cdot f(0)\, . $$ Then $g(-1) = g(0) = g(1)$. Repeated application of Rolle's theorem shows that $g'$ has two distinct zeros, and therefore $g''(c) = 0$ for some $c \in (-1, 1)$. So $$ 0 = g''(c) = f''(c) + 2f(0) \implies f''(c) = -2f(0) = 6 \, . $$

$\endgroup$
1
$\begingroup$

Note that by Taylor's theorem we have two numbers $c_1,c_2$ with $-1<c_1<0<c_2<1$ such that $$f(-1)=f(0)-f'(0)+\frac{f''(c_1)}{2},\,f(1)=f(0)+f'(0)+\frac{f''(c_2)}{2}$$ Adding these and noting that $f(0)=-3, f(-1)=f(1)=0$ we get $$\frac{f''(c_1)+f''(c_2)}{2}=6$$ It now follows via intermediate value property of derivatives that there is a $c\in[c_1,c_2]$ such that $f''(c) =6$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.