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If the curve ( equation in image) has its local extremum at x=d then 0 -3 image attached .

If the curve ( equation in image) has its local extremum at x=d then 
0
-3
image attached . 

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Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
Hello Student,
Please find answer to your question below

f(x) = x(x+3)e^{\frac{-x}{2}}
f(x) = (x^{2}+3x)e^{\frac{-x}{2}}
f'(x) = (x^{2}+3x)e^{\frac{-x}{2}}.\frac{-1}{2} + e^{\frac{-x}{2}}.(2x+3)
f'(x) = e^{\frac{-x}{2}}.(2x+3-\frac{3x}{2}-\frac{x^{2}}{2})
f'(x) = e^{\frac{-x}{2}}.(\frac{x}{2}+3-\frac{x^{2}}{2})
To find maxima & minima
f'(x) = e^{\frac{-x}{2}}.(\frac{x}{2}+3-\frac{x^{2}}{2}) = 0
e^{\frac{-x}{2}}.(\frac{x}{2}+3-\frac{x^{2}}{2}) = 0
x^{2}-x-6 = 0
x^{2}-3x+2x-6 = 0
(x-3)(x+2) = 0
x =3, -2
If you just need to find the point of minima, then
x = 3
is minima. Because
f(-3) = 0
f(2) = 10.e
So. x = -3 is the minima.

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