Derivatives and roots

Let's take a look at $f$ and $f'$ :

• $f(x) = {e}^{x} + \ln(x+1)-ax$

• $f'(x) = {e}^{x} + \frac{1}{x+1}-a$

Obviously $x$ has to be bigger than $-1$ . And it's easy to conclude that with $x \rightarrow -1$ , $f(x) \rightarrow -\infty$ ; $x \rightarrow +\infty$ , $f(x) \rightarrow +\infty$ . That means that $f'$ has positive values for these regions.

Considering $f''$ , we can notice that it's actually independent form $a$ :

• $f''(x) = {e}^{x} - \frac{1}{{(x+1)}^{2}}$

Other thing to say about $f''$ is that it is constantly increasing, meaning $f''(x) = 0$ only when $x=0$ . That mean $f'$ reaches it's minimum at $x=0$ . If $f'(0) > 0$ then there are no ${x}_{1}, {x}_{2}$ . Then:

• $f'(0) = {e}^{0} + \frac{1}{0+1}-a < 0$

• $1 + 1 < a$

• $a > 2$

Since ${x}_{1} < {x}_{2}$ and {f'(0) < 0} while $f'(0) > x$ for $x \rightarrow -1$ and $x \rightarrow +\infty$ , then ${x}_{1} < 0 < {x}_{2}$ but also $f({x}_{1}) > f(0) > f({x}_{2})$ . Which means $f({x}_{2}) - f({x}_{1}) < 0$ .

Thus, $2 \ln a > 2 \ln 2 > 0 > f({x}_{2}) - f({x}_{1})$