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Question:

Let f(x) = { x²|cos(πx)| ; x ≠ 0, 0 ; x = 0, then f is:

differentiable both at x=0 and at x=2

differentiable at x=0 but not differentiable at x=2

not differentiable at x=0 but differentiable at x=2

differentiable neither at x=0 nor at x=2

Solution:

Differentiability at x=0:
f'(0+) = limh→0 f(0+h) - f(0) / h
= limh→0 h²|cos(πh)| / h
= limh→0 h|cos(πh)| = 0
f'(0-) = limh→0 f(0-h) - f(0) / -h
= limh→0 h²|cos(-πh)| / -h
= limh→0 -h|cos(πh)| = 0
Since f'(0+) = f'(0-) = 0, f(x) is differentiable at x = 0.

Differentiability at x=2:
f(x) = x²|cos(πx)| = x²|cos(2π)| = x² for x near 2.
f'(x) = 2x
f'(2) = 4
Since f'(2) exists, f(x) is differentiable at x = 2.
Therefore, f(x) is differentiable both at x=0 and at x=2.