# Determining values where a function is not differentiable

Given

$$g(x) = \begin{cases} -1-2x & \text{if }x< -1,\\ x^2 & \text{if }-1\leq x\leq1,\\ x & \text{if }x>1, \end{cases}$$

determine at which values $g(x)$ is differentiable.

The approach I have taken with this question is to determine the values at which it is not differentiable, which will tell me all other values will be. I know that the function will not be differentiable where the limit at a given value does not exist. If I differentiate this function I get:

$$g'(x) = \begin{cases} -2 & \text{if }x< -1,\\ 2x & \text{if }-1\leq x\leq1,\\ 0 & \text{if }x>1. \end{cases}$$

I am a little bit lost as to how to proceed with this question - if I can show that the left hand and right hand limits disagree, then I can determine where the function is not differentiable, and therefore where it is differentiable. Am I heading in the right direction here?

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Yes, you're headed on the right path, I think. The function $g(x)$ is continuous, and its derivative exists everywhere except at $1$: $g'(1)$ does not exist since the left and right derivatives, which do exist, are different. Two nit-picks: $g'(x)=2x$ for $-1\le x\lt1$ (not $x\le1$), and $g'(x)=1$ (not zero) for $x\gt1$. –  bgins Jun 19 '12 at 21:30

First, note that $g'(x)=1$ for $x>1$.

On the interior of each of the intervals $(-\infty,-1)$, $[-1,1]$, and $(1,\infty)$ $g$ is differentiable since each component function is. The only question is what happens at the endpoints of these intervals.

At $x=-1$, the value of both component functions is $1$ and the derivative of both component functions is $2$. This means we get $$\lim_{h\to0^-}\frac{g(-1+h)-g(-1)}{h}=-2\tag{1}$$ Because $x^2=-1-2x$ at $x=-1$, we can use $g(x)=-1-2x$ for the computation of $(1)$.

Furthermore, we get $$\lim_{h\to0^+}\frac{g(-1+h)-g(-1)}{h}=-2\tag{2}$$ using $g(x)=x^2$ for the computation of $(2)$.

Since the derivatives computed in $(1)$ and $(2)$ are the same, we get that $g(x)$ is differentiable at $x=-1$.

At $x=1$, the value of both component functions is $1$, however, the derivative of $x^2$ is $2$ and the derivative of $x$ is $1$. This means that $$\lim_{h\to0^-}\frac{g(1+h)-g(1)}{h}=2\tag{3}$$ using $g(x)=x^2$ for the computation of $(3)$.

However, we get $$\lim_{h\to0^+}\frac{g(1+h)-g(1)}{h}=1\tag{4}$$ Because $x=x^2$ at $x=1$, we can use $g(x)=x$ for the computation of $(4)$.

Since the derivatives computed in $(3)$ and $(4)$ are different, $\lim\limits_{h\to0}\frac{g(1+h)-g(1)}{h}$ does not exist, and therefore $g(x)$ is not differentiable at $x=1$.

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