Given $f(x)=-2\arctan{\left(e^x\right)}+\ln\left(e^x+2\right)$ solve $f(x)=0$.

Question

I want to find the intersection with the $x$ axis of the following function:

$$f(x)=-2\arctan{\left(e^x\right)}+\ln\left(e^x+2\right)$$

I need to find

$$f(x)=0$$

So we have

$$-2\arctan\left(e^x\right)+\ln\left({e^x+2}\right)=0\rightarrow$$

$$\ln\left({e^x+2}\right)=2\arctan\left(e^x\right)$$

Because of the logarithim, I need to transform the right side into a logarithm as well so that I can eventually "remove" the logarithms and solve the equation, but I don't know how to do that. Any hints?

Answer 1

Write $t=e^x>0$:

$$f'(x) = {-2e^x\over 1+e^{2x}}+ {e^x\over e^{x}+2}=t{t^2-2t-3\over (t^2+1)(t+2)}$$

$$=t{(t-3)(t+1)\over (t^2+1)(t+2)}$$

so for $x\leq \ln 3$ it is decresing and for $x\geq \ln3$ it is increasing, so it has at most two solutions.

Answer 2

Hint: You will Need a numericval method, the two Solutions are given by $$x\approx -0.71157913296093670740$$ or $$x\approx 2.9343041578235969617$$

Answer 3

A numerical answer cones from the following steps.
1. Start with $y=e^x=2$.
2. Take ln
3. Divide by 2.
4. Take tangent
5. add 2
6. Go to step 2