Expert: Alon Mandes - 9/18/2008

Question
I'm having truble solving this problem

f(x)=|25x^2-16|

it asks to find the points in which the function is not differentiable. I think that I have to get the limits from both sides but am having truble with the piece wise function for it.

thanks
Anna

Answer
Ok Anna, the function f(x)=|25x^2-16| is defined :
f(x)=25x^2-61 if 25x^2-16>0 & -25x^2+16 if 25x^-16<0  OR
f(x)=25x^2-61 if x>4/5 & -25x^2+16 if x<4/5. So, the function won't
be differentiable in the points 4/5 & -4/5. Let's prove that by
calculating the derivative's limit at these points :

When x=4/5.
Calculating the Limit from the right :
Lim(h->0 +)  (1/h) *  |25(x+h)^2-16|-|25(x)^2-16| =
=Lim(h->0 +)  (1/h) *   25*[(4/5)+h]^2-16 - {25*(4/5)^2-16}
=Lim(h->0 +)  (1/h) *   25*[16/25+(8/5)h+h^2]-16 - {25*(16/25)-16}
=Lim(h->0 +)  (1/h) *   16+40h+25h^2-16-16+16
=Lim(h->0 +) (40h+25h^2)/h = 40.

Calculating the Limit from the left :
Lim(h->0 -)  (1/h) *  |25(x+h)^2-16|-|25(x)^2-16| =
=Lim(h->0 -)  (1/h) *   -25*[(4/5)+h]^2+16 - {-25*(4/5)^2+16}
=Lim(h->0 -)  (1/h) *   -25*[16/25+(8/5)h+h^2]+16-{-25*(16/25)+16}
=Lim(h->0 -)  (1/h) *   -16-40h-25h^2+16+16-16
=Lim(h->0 -) (-40h-25h^2)/h = -40.

So now we can see that Lim(h->0 +) IS NOt Equal Lim(h->0 -)
Which means the function is not differentiable at this point.

Anna, I will leave the other point -4/5 for you as an exercise,
just follow what I did above.
Any questions, I'm here.

Alon.