Let $F(x)=\frac{e^x-x-1}{x^2}$ for $x \neq 0$ and $F(x)=a$ for $x=0$.
Find $a$ so that $F$ is differentiable in $\mathbb{R}$ and then find $F$.
Show that $F$ (with the above value of $a$) is stricly increasing and stricly convex.
In order $F$ to be differentiable in $\mathbb{R}$, it has to be continous in $\mathbb{R}$.
It is continuous for $x \neq 0$. For $x=0$, it has to hold that $\lim_{x \to 0} F(x)=F(0)=a \Rightarrow \lim_{x \to 0} \frac{e^x-x-1}{x^2}=\lim_{x \to 0} \frac{e^x-1}{2x}=\lim_{x \to 0} \frac{e^x}{2}=a \Rightarrow a=\frac{1}{2}$.
For $x \neq 0$, $F'(x)=\frac{(e^x-x-1)' x^2 -(x^2)' (e^x-x-1)}{x^4}=\frac{(e^x-1) x^2-2x (e^x-x-1)}{x^4}=\frac{x^2 e^x-x^2-2xe^x+2x^2+2x}{x^4}=\frac{x^2 e^x+x^2-2xe^x+2x}{x^4}=\frac{xe^x (x-2)+x(x+2)} {x^4}$
For $x=0$, $F'(0)=\lim_{x \to 0} \frac{F(x)-F(0)}{x-0}=\lim_{x \to 0} \frac{\frac{e^x-x-1}{x^2}-\frac{1}{2}}{x}=\lim_{x\to 0} \frac{2e^x-2x-2-x^2}{2 x^3}=\lim_{x \to 0} \frac{2 e^x -2-2x}{6 x^2}=\lim_{x \to 0} \frac{2e^x-2}{12x}=\lim_{x\to 0} \frac{2 e^x}{12}=\frac{1}{6}$.
In order to show that $F$ is striclty increasing, we could show that $F'(x)>0$ for any $x$.
For $x>2$ this is obvously true.
But how can this be shown for $x<2$ ? Or do we show the strict monotony somehow else?
EDIT: In order to show that $F$ is also strictly convex, I have calculated $F''(x)=\frac{x^2 e^x-4xe^x-2x+6e^x-6}{x^4}$.
Then we set $w(x)=x^2 e^x-4xe^x-2x+6e^x-6$.
Then $w'(x)=xe^x(x−2)+2(e^x−1)$.
Then $w''(x)=x^2 e^x$.
But this implies that $w'(x)>0$ for $x>0$ and $w′(x)<0$ for $x<0$. Does this help?
We have $$F'(x)= \begin{cases} \dfrac{e^x(x-2)+x+2}{x^3} & x\ne 0 \\ \dfrac{1}{6} & x = 0 \end{cases} $$ Let $g(x)=e^x(x-2)+x+2$. We can show that $g(x)>0$ for $x>0$ and $g(x)<0$ for $x<0$ which would imply that $F'(x)>0$ for all $x$. $g'(x)=e^x(x-1)+1$ and $g''(x)=e^x x$. From the sign of $g''$ we see that $g'$ is strictly decreasing in $(-\infty, 0]$ and strictly increasing in $[0,\infty)$. Hence, $g'(x)\geq g'(0)=0$ (with equality only if $x=0$) which means that $g$ is a strictly increasing function. Then, for $x<0$, $g(x)<g(0)=0$ and for $x>0$, $g(x)>g(0)=0$.