Question

If $f\left(x\right)=x^{3/2}(3x-10),x\ge -0)$, then $f\left(x\right)$ is decreasing in

• A. $(-\infty ,0)\cup (0,\infty )$
• B. $(2,\infty )$
• C. $(-\infty ,-1]\cup [1,\infty )$
• D. $(-\infty ,0]\cup [2,\infty )$

Answer 1

Answer: (B)

$(2,\infty )$

R.E.F image

$f^{\prime }\left(x\right)=3x^{3/2}+(3x-10)\frac{3}{2}{x}^{1/2}$

Putting $f^{\prime }\left(x\right)=0\Rightarrow 3x^{3/2}+(3x-10)\frac{3}{2}{x}^{1/2}=0$

$\Rightarrow 2x^{3/2}=(10-3x)x^{1/2}$

$\Rightarrow 2x=10-3x$

$\Rightarrow 5x=10$

$\Rightarrow 5x=10$

$\Rightarrow x=2$

value of $x$ Intervals sign of $f^{\prime }\left(x\right)$

$x>2$ $xϵ(2,\infty )$ $<0$ decreasing

$x<2$ $xϵ(-\infty ,2)$ $>0$ Increasing