← Back to learning objective

Differentiability at x = 0 for a piecewise power function

Consider the piecewise function f(x)={x3if x<0x2if x≥0f(x) = \begin{cases} x^3 & \text{if } x < 0 \\ x^2 & \text{if } x \ge 0 \end{cases}

Using the limit definition of the derivative, determine whether f(x)f(x) is differentiable at x=0x = 0. Compute the left-hand limit lim⁡h→0−f(0+h)−f(0)h\displaystyle\lim_{h \to 0^-} \frac{f(0+h)-f(0)}{h} and the right-hand limit lim⁡h→0+f(0+h)−f(0)h\displaystyle\lim_{h \to 0^+} \frac{f(0+h)-f(0)}{h}, and use these to state your conclusion about differentiability at x=0x=0.

Differentiability at x = 0 for a piecewise power function graph
Hint

Since f(0)=02=0f(0) = 0^2 = 0, use the definition f′(0)=lim⁡h→0f(h)−0hf'(0) = \lim_{h\to 0} \frac{f(h) - 0}{h}. Split into cases h<0h<0 (use x3x^3) and h>0h>0 (use x2x^2), simplify each fraction before taking the limit, and compare the two one-sided results.

Step-by-step solution

First note f(0)=02=0f(0) = 0^2 = 0. For h→0−h \to 0^-, we have h<0h<0 so f(h)=h3f(h) = h^3, giving f(h)−f(0)h=h3h=h2\frac{f(h)-f(0)}{h} = \frac{h^3}{h} = h^2. Taking the limit as h→0−h \to 0^- gives lim⁡h→0−h2=0\lim_{h\to 0^-} h^2 = 0.

For h→0+h \to 0^+, we have h≥0h \ge 0 so f(h)=h2f(h) = h^2, giving f(h)−f(0)h=h2h=h\frac{f(h)-f(0)}{h} = \frac{h^2}{h} = h. Taking the limit as h→0+h\to 0^+ gives lim⁡h→0+h=0\lim_{h\to 0^+} h = 0. Since both one-sided limits equal 00, the two-sided limit exists and equals 00, so ff is differentiable at x=0x=0 with f′(0)=0f'(0) = 0.

Related problems

Practice similar Calculus problems interactively.

Start Closing the Gap