Vtyjkyuk

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

$$f(x)=-2arctanleft(e^xright)+lnleft(e^x+2right)$$

I need to find

$$f(x)=0$$

So we have

$$-2arctanleft(e^xright)+lnleft(e^x+2right)=0rightarrow$$

$$lnleft(e^x+2right)=2arctanleft(e^xright)$$

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?

Write $t=e^x>0$:

$$f'(x) = -2e^xover 1+e^2x+ e^xover e^x+2=tt^2-2t-3over (t^2+1)(t+2)$$

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

so for $xleq ln 3$ it is decresing and for $xgeq ln3$ it is increasing, so it has at most two solutions.

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

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