Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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

1 Answer 1

up vote 2 down vote accepted

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 $x=1$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.