alleryFor f(x) = x² ln x on (0, ∞), find x where f has local minimum:
Ax = 1
Bx = e
Cx = 0
Dx = 1/sqrt(e) (since f'(x) = 2x ln x + x = 0 ⇒ ln x = -1/2)
Answer & Solution
Correct answer: D. x = 1/sqrt(e) (since f'(x) = 2x ln x + x = 0 ⇒ ln x = -1/2)
f'(x) = 2x ln x + x = x(2 ln x + 1) = 0. Since x > 0: 2 ln x + 1 = 0
We have → ln x = -1/2 → x.
That is → ln x = e^(-1/2).
So → ln x = 1/√e.
f''(x) > 0 there, so it's a local min.
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