Question

Find the intervals in which the function $f\left(x\right)=\frac{x^4}{4}-x^3-5x^2+24x+12$ is
(a) strictly increasing, (b) strictly decreasing.

Answer 1

$Wehave,f\left(x\right)=\frac{x^4}{4}-x^3-5x^2+24x+12\Rightarrow f^{\prime }\left(x\right)=x^3-3x^2-10x+24\Rightarrow f^{\prime \prime }\left(x\right)=3x^2-6x-10If{f}^{\prime \prime }\left(x\right)>0thefunctionisstrictlydecreasingandif{f}^{\prime \prime }\left(x\right)<0\Rightarrow 3x^2-6x-10=0\Rightarrow x=\frac{6\pm \sqrt{36+4\times 3\times 10}}{6}=\frac{6\pm \sqrt{156}}{6}=\frac{6\pm 12.4899}{6}\Rightarrow x=3.08or-1.08Wegetthat{f}^{\prime \prime }\left(x\right)<0when-1.08<x<3.08and{f}^{\prime \prime }\left(x\right)>0whenx>3.08andx<-1.08\therefore f\left(x\right)isstrictlyincreasingwhen-1.08<x<3.08andf\left(x\right)isstrictlydecreasingwhenx>3.08andx<-1.08$