As I need to prove that the h function "$h\left(x\right)=x^2-5x+\ln \left(\left|-x+4\right|\right)$" has one real root in $\left[0,2\right]$ i have tried to do this but i would like to know if is it correct or what i should improve.
Domain: $\mathbb{R}-\left\{4\right\}$, so the function is continuous in $\left[0,2\right]$
$h\left(2\right)=-5,3$ $\space$ and $\space$ $h\left(3\right)=-5,3$
By Bolzano Theorem, being $h\left(x\right)$ continue and $h\left(a\right)\cdot h\left(b\right)<0$ $\space$, at least the function admits one root.
Now to be sure that $h\left(x\right)$ has only one root at that interval, we have to study the sign of $h'\left(x\right)$
$h'\left(x\right)=\frac{-2x^2+13x-21}{-x+4}$
Sign of $h'\left(x\right)$ = image on https://i.stack.imgur.com/nJ2IH.jpg
As the sign of $h'\left(x\right)$ is decreasing monotone, we can affirm that the funcion has only one real root on $\left[0,2\right]$
I would just use the Intermediate Value Theorem to show that $h(x)=0$ for some $x\in[0,2]$ because $h(0)=\ln(4) > 0$ and since $4-e\in[0,2]$ then $$h(4-e) = (4-e)^2 - 5(4-3)+\ln(|4-4+e| )= (4-e)^2-5(4-e)+1$$ To see that $h(4-e) < 0$ notice that $(4-e)^2-5(4-e) = (4-e)(4-e-5) = -(4-e)(e+1)$ and determine that $(4-e)(e+1)>0$. Since there exists points $x_1, x_2\in[0,2]$ such that $h(x_1)<0<h(x_2)$ then the Intermediate Value Theorem tells of the existence of a root.
To differentiate, we notice on our domain that $h(x) = x^2-5x + \ln(4-x)$ and so $h'(x) = 2x - 5 - \frac{1}{4-x}$. As $2x-5 \in[-5,-1]$ for all $x\in[0,2]$ and $-\frac{1}{4-x}\in[-\frac{1}{2}, -\frac{1}{4}]$. Thus $h'(x)<0$ for $x\in[0,2]$. We use that $h$ is strictly decreasing and thus $h$ only has one root.