Find the left-hand and right-hand derivatives of f(x) =

Find the left-hand and right-hand derivatives of $f(x) = |x|$ at $x = 0$. Does $f'(0)$ exist?

Step 1: Write the difference quotient at $x = 0$

Since $f(0) = |0| = 0$: $\frac{\Delta y}{\Delta x} = \frac{f(0+\Delta x)-f(0)}{\Delta x} = \frac{|\Delta x|}{\Delta x}$.

Step 2: Compute the right-hand derivative

For $\Delta x \gt 0$ we have $|\Delta x| = \Delta x$, hence

\[ f'_{+}(0) = \lim_{\Delta x \to 0^{+}} \frac{|\Delta x|}{\Delta x} = \lim_{\Delta x \to 0^{+}} \frac{\Delta x}{\Delta x} = 1. \]

Step 3: Compute the left-hand derivative

For $\Delta x \lt 0$ we have $|\Delta x| = -\Delta x$, hence

\[ f'_{-}(0) = \lim_{\Delta x \to 0^{-}} \frac{|\Delta x|}{\Delta x} = \lim_{\Delta x \to 0^{-}} \frac{-\Delta x}{\Delta x} = -1. \]

Step 4: Decide whether $f'(0)$ exists

Both one-sided derivatives exist and are finite, but they are different: $f'_{+}(0) = 1 \neq -1 = f'_{-}(0)$.

A two-sided limit exists only when the two one-sided limits agree, so

\[ f'(0) = \lim_{\Delta x \to 0}\frac{|\Delta x|}{\Delta x} \quad\text{does not exist.} \]

Step 5: Verification

\[x \gt 0:\ f(x) = x \Rightarrow f'(x) = 1; \qquad x \lt 0:\ f(x) = -x \Rightarrow f'(x) = -1 \;\;\checkmark\]

The two branches of the graph are straight lines of slopes $-1$ and $+1$ meeting at the origin, so the one-sided derivatives computed above are exactly the slopes of those lines.