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  • Find the intervals in which the function <img alt="f\left ( x \right )= \frac{x^{4}}{4}-x^{3}-5x^{2}+24x+12" src="https://entrancecorner.codecogs.com/gif.latex?f%5Cleft%20%28%20x%20%5Cright%20%29%3D%20%5Cfrac%7Bx%5E%7B4%7D%7D%7B4%7D-x%5E%7B3%7D-5x%5E%7

Find the intervals in which the function f\left ( x \right )= \frac{x^{4}}{4}-x^{3}-5x^{2}+24x+12 is
(a) strictly increasing, (b) strictly decreasing.

 

 

 

 
 
 
 
 

Answers (1)

given function is:
f\left ( x \right )= \frac{x^{4}}{4}-x^{3}-5x^{2}+24x+12
\Rightarrow {f}'\left ( x \right )= \frac{4x^{3}}{4}-3x^{2}-10x+24
\Rightarrow {f}'\left ( x \right )=x^{3}-3x^{2}-10x+24
For initial points, put {f}'\left ( x \right )=0
\therefore x^{3}-3x^{2}-10x+24= 0
\Rightarrow \left ( x-2 \right )\left ( x^{2} -x-12\right )= 0
\Rightarrow\left ( x-2 \right )\left ( x-4 \right )\left ( x+3 \right )= 0
\Rightarrow x= 2,4,-3
Therefore, we have intervals \left ( -\infty ,-3 \right ),\left ( -3,2 \right ),\left ( 2,4 \right ) and \left ( 4,\infty \right )


                                             
 \because {f}'\left ( x \right )> 0\; in\left ( -3,2 \right )\cup \left ( 4,0 \right )
\therefore f\left ( x \right ) is strictly increasing in the interval \left ( -3,2 \right )\cup \left ( 4,0 \right ) and {f}'\left ( x \right )< 0 \: in\: \left ( -\infty ,-3 \right )\cup \left ( 2,4 \right )
\therefore f\left ( x \right ) is strictly decreasing in \left ( -\infty ,-3 \right )\cup \left ( 2,4 \right )
 

Posted by

Ravindra Pindel

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