Gradient difference 1

The derivatives are $f'(x)=4x^3+6x^2-6x$ and $g'(x)=-9x^2+18x-2$ . Let $h(x)=f'(x)-g'(x)=4x^3+15x^2-24x+2$ . To maximize $h(x)$ , we need to consider $h'(x)=0$ . The derivative of $h$ is $h'(x)=12x^2+30x-24$ . Solving $h'(x)=0$ yields two solutions: $x_1 \approx -3.1375$ and $x_2 \approx 0.63746$ . Furthermore, we have $h''(x)=24x+30$ so that $h''(x_1)<0$ , which means that $h$ is maximal at $x_1$ . Hence, the value that we are looking for is $h(x_1) \approx 101.4$ .