#### Please solve RD Sharma class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xxvix maths textbook solution

$\text { Increasing interval }(-3,2) \cup(4, \infty) \\ \text { Decreasing interval }(-\infty ,-3) \cup(2, 4)$

Given:

Here given that

$f(x)=\frac{x^{4}}{4}-x^{3}-5x^{2}+24x+12$

To find:

We have to find the intervals in which function is increasing and decreasing.

Hint:

Put f ‘(x) = 0 to find critical points and then use increasing and decreasing property.

Solution:

We have,

$f(x)=\frac{x^{4}}{4}-x^{3}-5x^{2}+24x+12$

Differentiating w.r.t. x, we get,

\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(\frac{x^{4}}{4}-x^{3}-5 x^{2}+24 x+12\right) \\ &=\frac{4 x^{3}}{4}-3 x^{2}-10 x+24 \\ &=x^{3}-3 x^{2}-10 x+24 \end{aligned}

At critical points,

\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow x^{3}-3 x^{2}-10 x+24=0 \\ &\Rightarrow x^{3}-2 x^{2}-x^{2}+2 x-12 x+24=0 \\ &\Rightarrow x^{2}(x-2)-x(x-2)-12(x-2)=0 \\ &\Rightarrow(x-2)\left(x^{2}-x-12\right)=0 \\ &\Rightarrow(x-2)\left(x^{2}-4 x+3 x-12\right)=0 \\ &\Rightarrow(x-2)(x-4)(x+3)=0\\ &\Rightarrow x=2;\: x=4;\: x=-3 \\ &\Rightarrow x=-3,2,4 \end{aligned}

The points x=-3, 2, 4 divide the real line into four disjoint intervals,

\begin{aligned} &(-\infty,-3),(-3,2),(2,4) \text { and }(4, \infty) \\ &\text { In intervals }(-3,2) \text { and }(4, \infty), f^{\prime}(x)>0 \end{aligned}

$\text { Clearly, } f^{\prime}(x)>0 \text { if }-3

\begin{aligned} &\text { In the intervals }(-\infty,-3) \text { and }(2,4) f^{\prime}(x)<0\\ &\text { Clearly, } f^{\prime}(x)<0 \text { if }-\infty

$\text { Thus, } f(x) \text { is increasing in the intervals }(-3,2) \cup(4, \infty) \text { and } \text { decreasing in the intervals }(-\infty,-3) \cup(2,4) \text { . }$