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At where f(x)=|(x^2)-9| is differentiable?

Answer 1

Everywhere except x=3 and x=-3

Since #x^2-9# is a polynom it is differentiable everywhere. Then, in first approximation #|x^2-9|# is differentiable everywhere except when #x^2-9=0#, which is #x=+-3#. Now we must calculate derivates at the points #-3^+, -3^-# and #3^+, 3^-#, and see if the function derivate is continuous.

If #x^2-9>0#, #f(x)=x^2-9#, and #f'(x)=2x#

If #x^2-9<0#, #f(x)=-x^2+9#, and #f'(x)=-2x#

When #x=3#:

#f'(3^-)=-2*3=-6# #f'(3^+)=2*3=6# The values are different, therefore f(x) is not differentiable in x=3.

The same conclusion could be obtained when x=-3.

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Answer 2

Emily Ammerman

The function ( f(x) = |(x^2) - 9| ) is differentiable at all points except at ( x = -3 ), ( x = 3 ), and where ( (x^2) - 9 = 0 ). Therefore, it is differentiable for all ( x ) such that ( x \neq -3 ), ( x \neq 3 ), and ( x \neq -3 ).

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Answer from HIX Tutor

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