# Proving second value of second derivative from function and first derivative

$$Let \ f: [-1,1] \rightarrow \mathbb{R}$$ 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!

## 4 Answers

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 $$[0,1]$$.

• 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] ? – Matt_G Oct 8 '18 at 5:38
• 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. – Eric Wofsey Oct 8 '18 at 5:44

$$-3=\frac{f(-1)-f(0)}{-1\ -0}=f'(a)$$ 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 $$0.Therefore $$f''(c)>3$$

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 $$0 = g''(c) = f''(c) + 2f(0) \implies f''(c) = -2f(0) = 6 \, .$$

Note that by Taylor's theorem we have two numbers $$c_1,c_2$$ with $$-1 such that $$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$$.