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I'm trying to solve this question:

Suppose $f''(x)\gt 0$ in $(a,+\infty)$ and there is $x_0\gt a$ such that $f'(x_0)\gt 0$. Prove that $\lim_{x\to +\infty}f(x)=+\infty$.

My attempt:

I know that there is $\alpha \in (a,x_0)$ such that $f''(\alpha)=\frac{f'(x_0)-f'(a)}{x_0 - a}\gt0$, then $f'(x_0)\gt f'(a).$

I don't know what to do with this information, I can't follow from that, I feel that I'm close to the solution, I need help.

Thanks a lot

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Is $f(\bullet)$ twice continuously differentiable? –  Inquest Feb 8 '13 at 6:34
1  
It doesn't need to be. Twice differentiable is all that is needed. –  copper.hat Feb 8 '13 at 6:35

5 Answers 5

up vote 4 down vote accepted

Hints:

  • For every $x\geqslant x_0$, $f'(x)\geqslant f'(x_0)$.
  • For every $x\geqslant x_0$, $f(x)\geqslant f(x_0)+f'(x_0)(x-x_0)$.
  • $\lim\limits_{x\to+\infty}f(x_0)+f'(x_0)(x-x_0)=+\infty$.
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Thank you very much for your answer, it was very helpful! –  user42912 Feb 8 '13 at 6:46

If $f''(x) >0$ on $I=[x_0, \infty)$, then $f'(x)$ is strictly increasing on $I$. In particular, $f'(x) \geq f'(x_0)$.

Since $f(x) = f(x_0)+\int_{x_0}^x f'(t) dt$, we have $f(x) \geq f(x_0)+\int_{x_0}^x f'(x_0) dt = f(x_0)+f'(x_0)(x-x_0)$. Since $f'(x_0)>0$, we see that $\lim_{x \to \infty} f(x) = \infty$.

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You know that $f'$ is increasing and that $f'(x_0) = m > 0$. Prove that $f(x) > m(x-x_0) + f(x_0)$ for all $x>x_0$.

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We have $f''(x)>0$ for all $x\in(a,\infty)$, so that $f'(x)>f'(x_0)>0$ for all $x\in (x_0,\infty)$. Thus $f(x)=f(x_0)+\frac{f(x)-f(x_0)}{x-x_0}(x-x_0)=f(x_0)+f'(b)(x-x_0)$ (for some $b>x_0$ by mean value theorem) $>f(x_0)+f'(x_0)(x-x_0)\to\infty$ as $x\to\infty$.

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Let's denote $f'(x_0)$ by $\beta$. First, prove that $f'(x) \ge \beta$ for all $x \ge x_0$. Then prove that $f(x) \ge f(x_0) + \beta (x - x_0)$ for all $x \ge x_0$, and you are done. Each of the two steps can be done using Lagrange's theorem.

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