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The function $f:(0,\infty)\rightarrow \mathbb{R}$ is twice differentiable in its domain and satisfies the following:

i) $f(2)=-1,$

ii) $f'(2)>0,$

iii) $f''(x)\geq 0, \ \text{in} \ [2,\infty).$

Show that:

a) $\lim_{x\rightarrow \infty}f(x)=\infty.$

b) $f$ has at most one root in $(2,\infty).$

c) $f$ has at least one root in $(2,\infty).$


Attempt:

a) We know that $f''\geq 0 $ in $[0,\infty)\Longrightarrow f'$ is strictly increasing in $(2,\infty)$ and that $f$ is convex in the same interval.

This implies that $f(x)\rightarrow\infty$ as $x\rightarrow\infty$ and a) is shown.

b)+c)

Since the above holds and since $f(2)=-1<0,$ we know that there exists a $c\in\mathbb{R}$ such that $f(c)>0.$ Thus, according to the intermediate value theorem there exists at least one $\xi\in(2,c)$ such that $f(\xi)=0$

Is this "proof" (it's a proof to me until i'm disproven) correct or am I really off the track here?

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  • $\begingroup$ a) : Consider the function $f \colon x \mapsto 1/x$. $f'$ is strictly increasing in $(2,\infty)$ but $f$ does not tend to $\infty$. So your argument is incomplete. $\endgroup$ – tristan Dec 20 '17 at 19:14
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    $\begingroup$ To be more precise with part (a), use MVT to get the bound $f'(x)\geq f'(2)$ for $x >2$. Then use MVT again to get $f(x) \geq f(2) + (x-2)f'(2)$. $\endgroup$ – Luke Peachey Dec 20 '17 at 19:19
  • $\begingroup$ @tristan. True! How would you suggest I fill in the gaps then? $\endgroup$ – Parseval Dec 20 '17 at 19:56
  • $\begingroup$ @LukePeachey: I'll try this. I'll get back if I get stuck. $\endgroup$ – Parseval Dec 20 '17 at 19:56
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1) $f''(x) \ge 0 , x \in [2,\infty)$ implies

$f'(x) $ is non decreasing.

Since $f'(2) >0$ , $f'(x) \ge f'(2) >0$ ,

$x \in [2,\infty).$

MVT:

$f(x)-f(2) = (x-2)f'(t)\ge$

$(x-2)f'(2)$, $2<t<x.$

$\lim_{x \rightarrow \infty}f(x) \ge$

$\lim_{ x\rightarrow \infty} [f(2) + (x-2)f'(2)] = \infty.$

b)c).

b) $f(2) = -1$.

Since $ \lim_{x \rightarrow \infty}f(x)=\infty$,

and $f$ continuos,

there is a $s$ with $f(s) =0$, $s \in [2,\infty)$ (IVT).

c) Since $f'(x) >0$,

$f(x)$ is strictly increasing, there is at most one $0$.

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You wrote

a) We know that $f''\geq 0 $ in $[0,\infty)\Longrightarrow f'$ is strictly increasing in $(2,\infty)$ and that $f$ is convex in the same interval.

This is false. For instance, if $f'(x) = 1$ for all $x \in (2, \infty)$, then $f''(x) = 0 \ge 0$ for all $x$...but $f'$ is not strictly increasing. There's a real difference between "less than" and "less than or equal to".

You're certainly on the right track; you just need to be more careful.

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