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Here's my question:

Let $f$ be a function which has a derivative in $\Bbb R$ such that $f'(x)\geq0$ and $f''(x)\geq0$ for all $x \in \Bbb R$

Prove that if there is some $a \in \Bbb R$ such that $f'(a)>0$ then $\lim_\limits{x\to\infty}=\infty$.

I have a hint in the question: Mean value theorem

What I know:

  • The function is monotonic increasing, so is the derivative. However I don't really know what to do with it.

I want to prove that the function is not bounded, and conclude it approaches infinity. Couldn't figure out how.

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The condition on the second derivative implies that the derivative $f'$ is nondecreasing - in particular, this means that

$$f'(x) \ge f'(a) > 0$$

for all $x \ge a$. Now one can conclude that $f$ must lie on or above the line

$$y = f(a) + f'(a) (x - a)$$

Formalize this with the mean value theorem: If $f$ takes a value below this line, conclude something about the derivative being too small.

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  • $\begingroup$ Can you elaborate on the equation you wrote $y=f(a) +f'(a) (x-a)$? Thanks. $\endgroup$ – Alan Feb 3 '16 at 21:07
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    $\begingroup$ It's the equation of the tangent line to $f$ at a particular point. $\endgroup$ – user296602 Feb 3 '16 at 21:10
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The solution I finally assembled:

Since $f''(x) \geq 0$, $f'(x)$ is monotonic increasing (but not necessarily strictly increasing). Therefore, for all $x>a, f'(x)\geq f'(a)>0$, since $f'(a)>0$. Therefore we can say the $f$ is monotonic increasing (strictly) in the interval $(a,\infty)$.

In the interval $[a,x]$ has a derivative and is continuous in $(a,x)$, so from the Mean Value Theorem, there is a scalar $c \in [a,x] $ such that $f'(c)=\frac{f(x)-f(a)}{x-a} \Rightarrow f(x)=f'(c)*(x-a)+f(a)$. Since $f'(a)<f'(c)$, $f(x)\geq f'(a)*(x-a)+f(a)$.

$\lim_\limits{x \to \infty} f'(a)*(x-a)+f(a) = \infty$ (limits arithmetic).

$f(x)\geq f'(a)*(x-a)+f(a) \Rightarrow \lim_\limits{x \to \infty} f(x)=\infty $.

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