Question
Find the intervals in which the function $\text{f(x)}=\frac{\text{x}^4}{4}-\text{x}^3-5\text{x}^2+24\text{x}+12$ is (a) strictly increasing, (b) strictly decreasing.
✓
Answer
$\text{f(x)}=\frac{\text{x}^4}{4}-\text{x}^3-5\text{x}^2+24\text{x}+12$
$\text{f}'(\text{x})=\frac{4\text{x}^3}{4}-3\text{x}^2-10\text{x}+24$
$\text{f}'(\text{x})=0$
$\text{x}^3-3\text{x}^2-10\text{x}+24=0$
$(\text{x}-2)(\text{x}^2-\text{x}-12)=0$
$(\text{x}-2)(\text{x}-4)(\text{x}+3)=0$
increasing in interval $(-3,2)\cup(4,\infty)$
decreasing in interval $(-\infty,-3)\cup(2,4)$
$\text{f}'(\text{x})=\frac{4\text{x}^3}{4}-3\text{x}^2-10\text{x}+24$
$\text{f}'(\text{x})=0$
$\text{x}^3-3\text{x}^2-10\text{x}+24=0$
$(\text{x}-2)(\text{x}^2-\text{x}-12)=0$
$(\text{x}-2)(\text{x}-4)(\text{x}+3)=0$
increasing in interval $(-3,2)\cup(4,\infty)$
decreasing in interval $(-\infty,-3)\cup(2,4)$
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