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?