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Explain Solution for RD Sharma Maths Class 12 Chapter 17 Maxima and Minima Exercise Fill in the blanks Question 19 maths textbook solution        

Answers (1)

Answer:

              12

Hint:

For maxima or minima we must have {f}'\left ( x \right )=0

Given:

y=-x^{3}+3 x^{2}+9 x-27

Solution:  we have, y=-x^{3}+3 x^{2}+9 x-27

To find the slope we differentiate the function once

{y}'=-3x^{2}+6x+9

To find the extremum points we again differentiate at equate it to zero

{{y}'}'=-6x+6

{{y}'}'=0

-6x+6=0

x=1

Now to find whether at the critical points we find a maxima or minima, we use the second derivative test

{{{y}'}'}'=-6

{{{y}'}'}'\left ( 1 \right )=-6

{{{y}'}'}'\left ( 1 \right )< 0

Hence we find maxima at x=1  in the equation of the slope. Hence the max value of the slope is {y}'\left ( 1 \right )  which is

{y}'\left ( 1 \right )=-3+6+9

{y}'\left ( 1 \right )=12

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