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

$(2,3)$

Explanation for correct option

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

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

The critical point of $f\left(x\right)$ are the values of $x$ for which $f\text{'}\left(x\right)=0$.

$f\text{'}\left(x\right)=0$

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

$\Rightarrow x^2–5x+6=0$

$\Rightarrow (x–2)(x–3)=0$

$\Rightarrow x=2\mathrm{or}3$

It is clear from the figure, that the interval $(2,3)$ is negative which means $f\left(x\right)$ is strictly decreasing in this interval.

Hence, $\mathrm{Option}\left(\mathrm{A}\right)$ is correct.