piece wise CRIT POINTS?

Question

x+9...-8 x^2 +x.....-3 5x-4...2

Answer 1

find crit. points on each interval:

f'(x)=1 for -8<= x <= -3, so no critical points here
f'(x)=2x+1, -3<- x <= 2, 2x+1=0 then x= -1/2, which is in the interval -3<- x <= 2, so there is one critical point in this piece.
f'(x)=5 for 2

Answer 2

I assume you mean

f(x) = x+9, -8 <= x <= -3
= x^2 + x, -3 <= x <= 2
= 5x-4, 2 < x < 5

actually, you can't have x <= -3 for the first piece AND x >= -3 for the second piece, because the pieces can't overlap. You have to choose one piece or the other to obtain the value at x=3.

in any case,
f'(x) = 1, -8 <= x <= -3
= 2x + 1, -3 < x <= 2
= 5, 2 < x < 5

At x = -3, the left-hand derivative is equal to 1, but the right-hand derivative is -5 (obtained by plugging x=-3 into 2x+1, which gives the derivative as x -> 3 from the right). These don't match, so f is not differentiable at -3. Therefore it's a critical point.

Also, at x=-1/2, f'(x) = 0, so that is also a critical point. There are no others.