How to find numbers $k$ such that $kx - \ln(ex + 1-x) $ is positive on $(0,1]$?

Question

I want to find a condition on $k$ such that $g(x)= kx - \ln(ex + 1-x) > 0$, $x\in [0,1] $.

At zero the function is zero.

So, to find a condition on $k$ I use $g'(x) > 0$ i.e.

$$ k > \frac{e-1}{ex + 1 -x}$$

Now if I put $x=0.5$, then $k > \frac{2(e-1)}{e + 1 }$

let $k = 0.95$ as it satisfies the condition. But, I see that with this $k$, $g(0.5) <0$. I am making some silly mistake but not finding where.

Answer 1

Hint

In order $$g(x)= kx - \ln(ex + 1-x) > 0, x\in [0,1]$$ k must be such that for no $x$ in the interval $g(x)$ could be negative.

So, first look where is located the extremum (corresponding to $g'(x)=0$), use the second derivative test to know nif it is a maximum of a minimum. If it is a minimum, express the value of $g(x_*)$ (corresponding to $g'(x_*)=0$) and find the condition for $k$.

I am sure that you can take from here. I hope and wish that you understand the reasonning; if not, just post?