Where does class="AM"> fail to be
differentiable?
By definition, |h(x)|=h(x) if
h(x)>0, 0 if h(x)=0, and -h(x) if
h(x)<0.
h(x) is a parabola, opening up (leading
coefficient is positive). It factors as h(x)=(x+2)(x+1), so h(x)=0 at -1 and -2.
Checking points in the intervals class="AM"> we find that h(x)>0 on
and class="AM">
and h(x)<0 on
(-2,-1).
So |h(x)|=h(x) on class="AM"> and |h(x)|=-h(x) on
(-2,-1).
We need to find the critical points, which occur
when h'(x)=0, h'(x) fails to exist, or at the endpoints of the intervals. Since h(x) is
a polynomial, it is differentiable everywhere, so we need only check the endpoints of
the intervals.
As class="AM"> from the left, the derivative is 2x+3, and
h'(-2)=-1.
As from the
right, the derivative is -2x-3 (*-h(x)=*)
, and h'(-2)=1. Since the derivative from the left does not agree with the derivative
from the right, the function is not differentiable at
x=-2.
As
from the left, the derivative is -2x-3, and h'(-1)=-1.
As class="AM"> from the right, the derivative is 2x+3, and
h'(-1)=1. Since the derivative from the right disagrees with the derivative from the
left, the function is not differentiable at
x=-1.
Therefore, the function is not
differentiable at x=-2,x=-1.
** A look at
the graph shows that there are cusps at x=-1,x=-2**
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