Maximum of function abs Let function $f(x)=|2x^3-15x+m-5|+9x$ for $x\in\left[0,3\right]$ and $m\in R$. Given that $\max f(x) =60$ with $x\in\left[0,3\right]$, find $m$.
I know how to solve this kind of problem for $g(x)=|2x^3−15x+m−5|$. However, the $+9x$ is confusing me.

Case 1:
Let $2x^3-15x+m-5\gt0$ at the point where maximum occurs.
Then our $f(x)$ becomes $2x^3-6x+m-5$
note that $f(x)$ decreases for $(0,1)$ and then increases for $x\gt1$ so the maximum of $f(x)$ is at $x=3$(as $x\in\left[0,3\right]$)
Plugging in the $f(3)=60$ we get $m = 29$.
We can put in $x=3$ and $m = 29$  in $2x^3-6x+m-5$ and verify that it is positive.

Case 2 :
Let $2x^3-15x+m-5\lt0$ at the point where maximum occurs
Then our $f(x)$ becomes $-2x^3+24x-m+5$
This function increases in $(0,2)$ and then decreases for $x\gt2$ so maximum for $f(x)$ is at $x=2$
Putting $f(2)=60$ we get $m=-23$.
We can put in $x=2$ and $m = -23$  in $-2x^3+24x-m+5$ and verify that it is negative.
So we get $m= \left({29\ \ and\ -23}\right)$
