Mathematics

Mathematics

Problem

Consider the function f(x)={x2,x≤0,x,x>0.f(x)=\begin{cases} x^{2}, & x\le 0,\\ \sqrt{x}, & x>0. \end{cases} Identify the point(s) at which ff fails to be differentiable and state whether the failure corresponds to a corner, a cusp, or a vertical tangent.
Intuitive explanation

The key is to check differentiability at the boundary where the formula changes—here, x=0x = 0—by computing the left and right derivatives and seeing if they match. A corner occurs when both one-sided derivatives exist but differ; a cusp when they exist with opposite signs and infinite magnitude; a vertical tangent when they're both infinite. You'll need to carefully evaluate lim⁡h→0−f(h)−f(0)h\lim_{h \to 0^-} \frac{f(h) - f(0)}{h} and lim⁡h→0+f(h)−f(0)h\lim_{h \to 0^+} \frac{f(h) - f(0)}{h} to determine which case applies.

Intuitive explanation

The key is to check differentiability at the boundary where the formula changes—here, x=0x = 0—by computing the left and right derivatives and seeing if they match. A corner occurs when both one-sided derivatives exist but differ; a cusp when they exist with opposite signs and infinite magnitude; a vertical tangent when they're both infinite. You'll need to carefully evaluate lim⁡h→0−f(h)−f(0)h\lim_{h \to 0^-} \frac{f(h) - f(0)}{h} and lim⁡h→0+f(h)−f(0)h\lim_{h \to 0^+} \frac{f(h) - f(0)}{h} to determine which case applies.

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