Question

Function $f\left(x\right)=x^2(x-2{)}^2$ is

• A. Increasing in $(0,1)\cup (2,\infty )$
• B. Decreasing in $(0,1)\cup (2,\infty )$
• C. Increasing function
• D. Decreasing function

Answer 1

Answer: (A)

Increasing in $(0,1)\cup (2,\infty )$

$f\left(x\right)=(x^2-2x{)}^2$

Hence

$f^{\prime }\left(x\right)=2(x^2-2x)(2x-2)$

$>0$

Or

$4(x-1)(x^2-2x)>0$

Or

$4x(x-1)(x-2)>0$

Hence

$x>0$ $x<1$ and $x>2$

Thus the function is increasing in the interval
$(0,1)\cup (2,\infty )$