$f(x) = e^x - x^2/2 - x + \ln x$ Find the monotonicity using derivatives $f(x) = e^x - x^2/2 - x + \ln x$
The answer given by my book is that $f(x)$ is strictly increasing for $x\in(0,+οο)$.
But it's giving me trouble. I tried:
$f'(x) = e^x - x- 1 + 1/x$ but I have no idea how to prove $f'(x)>0$.
Any help is appreciated.
 A: Here are two possibilities.

*

*From Taylor series, we know that $e^x = 1 + x + x^2/2 +  \text{(other positive terms)}$. Subtracting $x + 1$ leaves $\text{positive terms} + \frac{1}{x}$, which is positive on $(0, \infty)$.


*We might also proceed naively from first principles. Note that $f'(1) = e + 1 > 0$, and $f''(x) = e^x - 1 - 1/x^2$. For $x > 1$, we have that $e^x > 2 > 1 + \frac{1}{x^2}$. Thus since $f''(x) > 0$ for $x > 1$ and $f'(1) > 0$, the derivative $f'(x)$ is increasing and positive on $(1,\infty)$. All that's left is to consider the interval $(0,1)$.
In this interval, since $\frac{1}{x} > x$ and $e^x > 1$, we get the desired inequality. Patching them together gives your answer. $\diamondsuit$
A: See from $x \in [0,1/2]$ $1/x>1+x$ and for $x \in (1/2, \infty)$ $e^x+1/x> 1+x$ so we are done.
A: $$f'(x)=e^x-x-1+\frac1{x}$$
$$\frac 1{x} >0$$
Also,you can find equation of tangent to $e^x$ at $x=0$
It's equation is $y=1+x$ you can easily find that this line is below the curve for all values of x.Thus,$e^x>1+x$
Thus,$f'(x)>0$
