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  • If x=1 is a critical point of a function  then Option: 1

If x=1 is a critical point of a function f(x) = (3x^2 + ax - 2-a)e^x then
Option: 1 x = 1 is a local maxima and x = \frac{-2}{3} is a local minima of f.
Option: 2 x = \frac{-2}{3} is a local maxima and x = 1  is a local minima of f.
Option: 3 x = \frac{-2}{3} and x = 1  is a local minima of f.
Option: 4 x = \frac{-2}{3} and x = 1  is a local maxima of f.

Answers (1)

best_answer

\\f(x) = ( 3x^2 +ax-2-a)e^x \\ \\f'(x) = ( 6x +a)e^x +( 3x^2 +ax-2-a)e^x \\ \\f'(x)|_{at x =1} = ( 6 +a)e^1 +( 3 +a-2-a)e^1 = 0\\ a + 7 = 0 \\ a = -7 \\ \\f'(x) = ( 3x^2 -x-2)e^x \\ \text{for minima and maxima f'(x) = 0 } \\ \\f'(x) = ( 3x^2 -x-2)e^x = 0 \\ 3x^2 -x-2 = 0 \\ 3x^2 -3x+2x-2 = 0 \\ 3x(x-1)+2(x-1) = 0 \\ (x-1)(3x+2) = 0 \\ x = 1 \ or \ x = \frac{-2}{3}

\\f''(x) = ( 6x -1)e^x +( 3x^2 -x-2)e^x \\ \\f''(x) |_{at x = 1}= 5e > 0\\ \rightarrow $local minima at x =1 $ \\ \\f''(x) |_{at x = -2/3}= -5e^{\frac{-2}{3}} < 0\\ \rightarrow $local maxima at x =-2/3 $

 

Posted by

himanshu.meshram

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