Question

The function $f(x)=5+36x+3x^2-2x^3$ is decreasing in the interval

• A. $(-2,3)$
• B. $(2,-3)$
• C. $(-2,-3)$
• D. $(2,3)$

Answer 1

Answer: (A)

Given , $f\left(x\right)=5+36x+3x^2=2x^3$
$\Rightarrow f^{\prime }\left(x\right)=36+6x-6x^2$
For increasing or decreasing; $f^{\prime }(x)=0$
$\Rightarrow 6+x-x^2=0\Rightarrow x^2-x-6=0$
$\Rightarrow \left(x-3\right)\left(x+2\right)=0\Rightarrow x=3,-2$
Intervals:
$-\infty <x<-2f^{\prime }\left(x\right)=\left(-ve\right)\left(-ve\right)=\left(+ve\right)$, Increasing $-2<x<0$
$f^{\prime }\left(x\right)=\left(-ve\right)\left(+ve\right)=\left(-ve\right),$ Decreasing $0<x<3$
$f^{\prime }\left(x\right)=\left(-ve\right)\left(+ve\right)=\left(-ve\right),$ Decreasing $3<x<\infty$
$f^{\prime }\left(x\right)=\left(+ve\right)\left(+ve\right)=\left(+ve\right)$ Increasing
$\therefore$ The interval in which $f(x)$ is decreasing is $(-2,3)$