2

options:
A) $|f(x)|>1 $

B) $|f(x)|<1 $

C) $|f′(x)|>1 $

D) $|f′(x)|<1$

attempt:

I first tried using integration. $−1\le f′′(x)\le 1$

integrating from $0$ to $x$, $−x\le f′(x)−f′(0)\le x$

Again integrating from $0$ to $x$, $\frac{−x^2}{2}\le f(x)−f(0)−f′(0)x\le\frac{x^2}{2}$

at $x=1$, $−\frac{1}{2}\le f′(0)\le\frac{1}{2}$

from the equation, $−x\le f′(x)−f(0)\le x$ and by substituting the max value of $f'(0)$, $-x+0.5\le f′(x)\le x+0.5$

But this doesn't give me the correct answer.

Ludolila
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Aditya Dev
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  • By Rolle's theorem there's $c \in [0,1]$ such that $f'(c)=0$ – mlainz Aug 16 '15 at 10:35
  • Also, it is easy to find counterexamples for A and B. $f(x)=2$, $f(x)=0$. – mlainz Aug 16 '15 at 10:39
  • @mlainz It is not sufficient that $c \in [0, 1]$, as the counter-example $f'(x) = x$ shows. You actually need $c \in (0, 1)$ [this also follows from Rolle's theorem]. – Dominik Aug 16 '15 at 10:45

1 Answers1

3

The first three answers are clearly false, as you can see from considering a constant function.

To see that the last answer is correct, observe that by Rolle's theorem there is a $\xi \in (0, 1)$ such that $f'(\xi) = 0$. By the mean value theorem, for every $x \in [0, 1]$ there is a $\xi' \in (0, 1)$ that satisfies $f'(x) = f'(\xi) + f''(\xi')(x - \xi)$. Therefore $|f'(x)| \le |x - \xi| < 1$ (note that $\xi$ is in the interior of $[0, 1]$).

Aditya Dev
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Dominik
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    Why you are so sure about Riemann-integrability of $f''$? –  Aug 16 '15 at 10:57
  • @EugenCovaci $f''$ has an antiderivative, isn't this enough to ensure Riemann-integrability? In any way, the proof can also be modified to circumvent this by using the mean-value-theorem. There always exists a $\xi' \in (0, 1)$ with $f'(x) = f''(\xi')(x - \xi)$ and the same estimate as in the proof holds. – Dominik Aug 16 '15 at 11:08
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    It is not enough: (Fundamental theorem of calculus I). If F : [a, b] → R is continuous on [a, b] and differentiable in (a, b) with F′ = f where f : [a, b] → R is Riemann integrable etc. You can see F' has antiderivative, still f is required to be Riemann integrable –  Aug 16 '15 at 11:12
  • @EugenCovaci Indeed, a counter-example can be found here. I've adjusted the proof. Thank you! – Dominik Aug 16 '15 at 11:15
  • can you explain why you used this method? I am a beginner in calculus. Please help me understand this. I know MVT and rolle's theorem – Aditya Dev Aug 16 '15 at 12:34
  • Essentially the constraint $|f''| \le c$ says that starting from any given point $x_0$, the function $f'$ can't rise or fall faster than a linear function with slope $c$. Searching for a point $x_0$ that minimizes the possible values leads to the equation $f'(x_0) = 0$. – Dominik Aug 16 '15 at 12:46