How many roots does this exponential equation have? I have the following course assignment: Calculate the number of roots of the following equation: $ e^x = 6 \cdot \ln(x^2 + 1) $
What I have done so far:
$ f(x) = e^x - 6 \cdot \ln(x^2 + 1) $
I will look for the roots of the derivative and check where the function crosses OX as the function is continuous (Darboux):
$ f'(x) = e^x - \frac{12x}{x^2+1} $
$f'(x) \gt 0, x \leq 0 $
$f'(1) \lt 0$
$f'(2) \gt 0$
$ f(0) = 1, f(1) \lt 0, f(2) \lt 0 $
And there's the function graph:

I am missing the part (0,1) . Can somebody give me a hint? I guess it will look as I have drawn, but I cannot prove it.

Since $f'(x)=\frac{e^x(x^2+1)-12x}{x^2+1}$, setting $g(x)=e^x(x^2+1)-12x$ gives us
$$g'(x)=e^x(x+1)^2-12,\ \ g''(x)=e^x(x+1)(x+3).$$
Since $g'(-3)\lt 0,g'(1)\lt 0,g'(2)\gt 0$, we can see that there is only one real number $\alpha\ (1\lt \alpha\lt 2)$ such that 
$$g'(x)\lt 0\ (x\lt \alpha),\ g'(\alpha)=0,\ g'(x)\gt 0\ (x\gt\alpha).$$
So, since $g(0)\gt 0,g(1)\lt 0,g(2)\gt 0$, we can see that there are only two real numbers $\beta,\gamma\ (0\lt\beta\lt 1\lt\gamma\lt 2)$ such that 
$$g(\beta)=g(\gamma)=0,$$
which means $f'(\beta)=f'(\gamma)=0$.
From what you've got, we know that the number of roots is three.
