Question

The function $f\left(x\right)=2x^3-15x^2+36x+6$ is strictly decreasing in the interval

• A. $(2,3)$
• B. $(-\infty ,2)$
• C. $(3,4)$
• D. $(-\infty ,3)\cup (4,\infty )$
• E. $(-\infty ,2)\cup (3,\infty )$

Answer 1

The correct option is A $(2,3)$

$f\left(x\right)=2x^3-15x^2+36x+6$

$f^{\prime }\left(x\right)=6x^2-30x+36$

$f^{\prime }\left(x\right)=0$

$6x^2-30x+36=0$

$\Rightarrow x^2-5x+6=0$

$\Rightarrow (x-2)(x-3)=0$

$\Rightarrow x=2$ or $x=3$

Since, from figure in interval $(2,3)$ is negative it means $f\left(x\right)$ is strictly decreasing in this interval.

Hence, A is the correct option.