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

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### 1 Answer

*Where does `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 `(-oo,-2),(-2,-1),(-1,oo)` we find that h(x)>0 on `(-oo,-2)` and `(-1,oo)` and h(x)<0 on (-2,-1).

So |h(x)|=h(x) on `(-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 `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 `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**