Given $f(x)=αe^{x-1}+β\ln x-1$ at which the $C_f$ is tangential to the axis $x'x$ at the point $A(1,0)$.
I) Show that $α=1$ and $β=-1$.
II) Find the monotony of $f$ and then, show that $f(x) \ge0, \forall x>0.$
III) Find the surface area which is enclosed by the the graph of $g(x)=xf(x), x>0$, the axis $x'x$ and the straight line $x=2.$
IV) Show that $$f'(x)>{f(x)\over x+1}, \forall x>1.$$
Personal work:
I) Given the point $A(1,0)$ we know that $f(1)=0$...
II) Since $α=1$ and $\beta =-1$, then $f(x)=e^{x-1}-lnx-1, x>0$
For $x>0: f'(x)=\cdots=e^{x-1}-{1\over x}$
$$f'(x)=0 \iff e^{x-1}-{1\over x}=0 \iff e^{x-1}={1\over x}\iff _{x\neq 0} xe^{x-1}=1 \iff x=1$$
$$f'(x)>0\iff\cdots \iff xe^{x-1}>1 \iff x>1$$ Therefore $f$ is strictly increasing at $(0,+\infty).$
III) $$g(x)=xe^{x-1}-xlnx-x, x>0$$
$$\int_1^2|g(x)| dx=\int_1^2 g(x) dx=\int_1^2 {(xe^{x-1}-xlnx-x)} dx=\cdots$$
IV) This is the one that I'm struggling the most. Should I multiply both sides with $x+1$, where $x>1$, and then reach to an inequality that is true based upon the previous questions?
We already know
$$f(x)=e^{x-1}-\ln x -1$$
so $$f'(x)=e^{x-1}- x^ {-1}$$
Let me denote $$k(x):=\frac{f(x)}{x+1}$$
We know that $k(1)=\frac{f(1)}2=0$ and $f'(1)=0$.
Now we consider $$h(x):=f'(x)-k(x)$$
Notice that
$$h'(x)=f''(x)-\frac{f'(x)}{x+1}+\frac{f(x)}{(x+1)^2}\\=e^{x-1}+\frac{1}{x^2}-\frac{e^{x-1}}{x+1}+\frac{1}{x(x+1)}+\frac{f(x)}{(x+1)^2}\\ =e^{x-1}\left(\frac{x}{x+1}\right)+\frac{1}{x^2}+\frac{1}{x(x+1)}+\frac{f(x)}{(x+1)^2} >0$$ since $e^{x-1},x,x+1,x^2,f(x)>0$ for $x>1>0$
Since $h(1)=0$ and $h'(x)>0$ for $x>1$, $$h(x)=f'(x)-\frac{f(x)}{x+1}>0$$ for $x>1$ which implies $$f'(x)>\frac{f(x)}{x+1}$$ for all $x>1$.