1. ## f'(x)

Let g(x)=1-lxl.
(a) Evaluate f'(2).
(b) Evaluate f'(-3).
(c) Does f'(0) exist?

For (c) I believe that f'(0) does not exist because it is not locally linear. However for (a) and (b) I do not know how to find f'(of anything). Can someone please explain how to find this and derivatives in general. THANK YOU!

2. First try to graph it so you can understand it better.
you'll see that f'(0) does not exist because it's a cusp.
then for all negative values their slope will be possitive and for negative values of x, they'll be negative.
So if i'm right, f'(2) = -1 ; f'(-3) = 1

3. first off it's $g(x)=1-x$ for $x>0$ and $g(x)=1+x$ for $x<0,$ now $g'(x)=-1$ for $x>0$ and $g'(x)=1$ for $x<0.$

4. Originally Posted by asweet1
Let g(x)=1-lxl.
(a) Evaluate f'(2).
(b) Evaluate f'(-3).
(c) Does f'(0) exist?

For (c) I believe that f'(0) does not exist because it is not locally linear. However for (a) and (b) I do not know how to find f'(of anything). Can someone please explain how to find this and derivatives in general. THANK YOU!
all you need to know in this case is that a derivative is the slope of a function.

you should be able to sketch a graph of the function $y = 1 - |x|$, and since each piece is linear, finding the slope is an easy process.

... finally, you should be able to answer all three questions by looking at the slope of the graph at x = 2, x = 3, and x = 0.

as far as "how" to find a derivative, here's a place to start ...

Pauls Online Notes : Calculus I - Derivatives