Let $k$ be a real number. $$f(x)=x^3-3x^2+6x+k$$ and let $g(x)$ be the inverse function of $f(x)$. The equation $$4f'(x)+12x-18=(f'\circ g)(x)$$ has a real root on $[0, 1]$. Find $m^2+M^2$ where $m$ is the minimum value of $k$ and $M$ is the maximum value of $k$.
My work:
$$f'(x)=3x^2-6x+6\\\implies 4f'(x)+12x-18=12x^2-12x+6$$
Since $(g\circ f)(x)=x$, plug $x=f(x)$ to the equation: $$12(f(x))^2-12f(x)+6=f'(x)$$ Now let $$h(x)=12(f(x))^2-12f(x)+6-f'(x)$$ then $h(0)h(1)\leq 0$ given that $h$ is continuous.
$f(0)=k$, $f(1)=k+4$ implies $h(0)=12k(k-1)$, $h(1)=12k^2+84k+147=3(2k+7)^2$.
Thus I have $m=-7/2$ and $M=1$.
But the correct answer is $m=-8$ and $M=1$.