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# The interval in which y = x2 e–x is increasing is (A) (– 8, 8) (B) (–2, 0) (C) (2, 8) (D) (0, 2

Latika Leekha
7 years ago
The given function is y = f(x) = x2e–x
Then f’(x) = 2xe-x – x2e-x
For finding the critical points, f’(x) = 0
Hence, we have xe-x (2-x) = 0
This gives x = 0 and 2 as the criticla points.
We now find out the second derivative.
y” = f”(x) = x2e–x - 2xe-x + 2e–x
Putting x = 0 we get the second derivative as 2 which is positive.
Hence this shows that at 0 the function attains the minimum.
Now e–x is always positive so we just need to check x(2-x).
Now we try to find the interval where this term is positive.
The term x(2-x) will be positive when either both x and (2-x) are positive or both are negative.
x is negative only when x < 0 but then (2-x) is positive and hence the term x(2-x) is negative which does not serve our purpose.
Now (2-x) is positive if x < 2.
Hence, (0,2) is the interval in which both have the same sign.
Hence, the function is increasing in the interval (0, 2).
Thanks & Regards
Latika Leekha