Question

Find the intervals for $f\left(x\right)=\frac{x^4}{12}-\frac{5x^3}{6}+3x^2+7$ in which it is concave upward

• A. $(-\infty ,0)\cup (3,\infty )$
• B. $(-\infty ,2)\cup (3,\infty )$
• C. $(-\infty ,-1)\cup (1,\infty )$
• D. None of these

Answer 1

The correct option is B $(-\infty ,2)\cup (3,\infty )$

Given, $f\left(x\right)=\frac{x^4}{12}-\frac{5x^3}{6}+3x^2+7$
$f^{\prime }\left(x\right)=\frac{x^3}{3}-\frac{5x^2}{2}+6x$
$f^{\prime \prime }\left(x\right)=x^2-5x+6=(x-2)(x-3)$
Now for $f\left(x\right)$ to be concave upward $f^{\prime \prime }\left(x\right)>0$

$\Rightarrow x\in (-\infty ,2)\cup (3,\infty )$