Compare the right-hand and left-hand derivatives to determine whether or not the function is differentiable at the point whose coordinates are given.
- $$y = \frac { 1 } { x } \quad y = - 1$$

A) Since $\lim _ { x \rightarrow 1 ^ { + } } f ^ { \prime } ( x ) = 0$ while $\lim _ { x \rightarrow 1 ^ { - } } f ^ { \prime } ( x ) = 1$, $f ( x )$ is not differentiable at $x = - 1$.
B) Since $\lim _ { x \rightarrow 1 ^ { + } } f ^ { \prime } ( x ) = - 1$ while $\lim _ { x \rightarrow 1 ^ { - } } f ^ { \prime } ( x ) = 0 , f ( x )$ is not differentiable at $x = - 1$.
C) Since $\lim _ { x \rightarrow 1 ^ { + } } f ^ { \prime } ( x ) = 0$ while $\lim _ { x \rightarrow 1 ^ { - } } f ^ { \prime } ( x ) = - 1 , f ( x )$ is not differentiable at $x = - 1$.
D) Since $\lim _ { x \rightarrow 1 ^ { + } } f ^ { \prime } ( x ) = 0$ while $\lim _ { x \rightarrow 1 ^ { - } } f ^ { \prime } ( x ) = 0$, $f ( x )$ is differentiable at $x = - 1$.

Question

Compare the right-hand and left-hand derivatives to determine whether or not the function is differentiable at the point whose coordinates are given.
- $$y = \frac { 1 } { x } \quad y = - 1$$

A) Since $\lim _ { x \rightarrow 1 ^ { + } } f ^ { \prime } ( x ) = 0$ while $\lim _ { x \rightarrow 1 ^ { - } } f ^ { \prime } ( x ) = 1$, $f ( x )$ is not differentiable at $x = - 1$. 
B) Since $\lim _ { x \rightarrow 1 ^ { + } } f ^ { \prime } ( x ) = - 1$ while $\lim _ { x \rightarrow 1 ^ { - } } f ^ { \prime } ( x ) = 0 , f ( x )$ is not differentiable at $x = - 1$. 
C) Since $\lim _ { x \rightarrow 1 ^ { + } } f ^ { \prime } ( x ) = 0$ while $\lim _ { x \rightarrow 1 ^ { - } } f ^ { \prime } ( x ) = - 1 , f ( x )$ is not differentiable at $x = - 1$. 
D) Since $\lim _ { x \rightarrow 1 ^ { + } } f ^ { \prime } ( x ) = 0$ while $\lim _ { x \rightarrow 1 ^ { - } } f ^ { \prime } ( x ) = 0$, $f ( x )$ is differentiable at $x = - 1$.

Quizplus · Accepted Answer

The function is given by $y= \frac{1}{x}$ which is a hyperbola with vertical and horizontal asymptotes at $x=0$ and $y=0$, respectively. Also, $y=-1$ is a horizontal line.
The point of interest is where the hyperbola intersects the horizontal line, which occurs at $x=-1$. Note that this point is not defined in the original function since $\frac{1}{x}$ is not defined at $x=0$. However, we can still talk about the limit of the slope of the tangent line at $x=-1$ from the left and from the right.

From the left,
$$f'(x) = \lim_{h 	o 0^-} \frac{f(x+h)-f(x)}{h} = \lim_{h 	o 0^-} \frac{\frac{1}{x+h} - \frac{1}{x}}{h} = \lim_{h 	o 0^-} \frac{-1}{x(x+h)h} = \frac{-1}{x^2}$$
Hence, 
$$\lim_{x 	o -1^-} f'(x) = \lim_{x 	o -1^-} \frac{-1}{x^2} = -1$$

From the right,
$$f'(x) = \lim_{h 	o 0^+} \frac{f(x+h)-f(x)}{h} = \lim_{h 	o 0^+} \frac{\frac{1}{x+h} - \frac{1}{x}}{h} = \lim_{h 	o 0^+} \frac{-1}{x(x+h)h} = \frac{-1}{x^2}$$
Hence,
$$\lim_{x 	o -1^+} f'(x) = \lim_{x 	o -1^+} \frac{-1}{x^2} = -1$$

Since the two one-sided derivatives exist and are equal, $f(x)$ is differentiable at $x=-1$. But note that $f(x)$ is not continuous at $x=0$ and hence not differentiable at $x=0$.

Compare the Right-Hand and Left-Hand Derivatives to Determine Whether or Not