where does h(x)=|x^2+3x+2| fail to be differentiable?

Where does class="AM">`h(x)=|x^2+3x+2|` 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">`(-oo,-2),(-2,-1),(-1,oo)` we find that h(x)>0 on
`(-oo,-2)` and class="AM">`(-1,oo)` and h(x)<0 on
(-2,-1).

So |h(x)|=h(x) on class="AM">`(-oo,-2),(-1,oo)` 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">`x->-2` from the left, the derivative is 2x+3, and
h'(-2)=-1.
As `x->-2` from the
right, the derivative is -2x-3 (*-h(x)=`-x^2-3x-2`*)
, 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 `x->-1`
from the left, the derivative is -2x-3, and h'(-1)=-1.
As class="AM">`x->-1` 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** src="/jax/includes/tinymce/jscripts/tiny_mce/plugins/asciisvg/js/d.svg"
sscr="-7.5,7.5,-5,5,1,1,1,1,1,300,200,func,abs(x^2+3x+2),null,0,0,,,black,1,none,func,x^2+3x+2,null,0,0,-7,-2,black,1,none"/>