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  • Please solve RD Sharma class 12 Chapter maxima and minima exercise 17.5 question 35 maths textbook solution.

Please solve RD Sharma class 12 Chapter maxima and minima exercise 17.5 question 35 maths textbook solution.

Answers (1)

5

Hint: For maximum or minimum value of S, we must have \frac{dS}{dx}=0

Given:

y= -x^3+3x^2+2x-27                 ......(1)

Slope = \frac{dy}{dx}=-3x^2+6x+2

Solution:

Now,

m=-3x^2+6x+2

\frac{dm}{dx}=-6x+6

\frac{dm}{dx}=0

-6x+6=0

x = 1

Substituting the value of x in eqn(1)

y =(1)^3+3(1)^2+2(1)-27

=-23

\frac{d^2m}{dx^2}=-6<0

So the slope is maximum  x= 1, y = -23 at  (1,-23)

maximum slope = -3(1)^2+6(1)+2

= 5

 

Posted by

infoexpert24

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