Q 19. - Calculus

Question

Suppose f is a piecewise-defined function, equal to g(x) if x < 2 and h(x) if x ≥ 2, where g and h are continuous and differentiable everywhere. If g(2) = h(2), is the function f necessarily differentiable at x = 2? Why or why not?

Step-by-Step Solution

Answer

$\mathrm{The} \mathrm{function} \mathrm{is} \mathrm{continuous} \mathrm{at} \mathrm{x}=2 \mathrm{but} \mathrm{not} \mathrm{differentiable} \mathrm{at} \mathrm{x}=2.$

Step 1. Given information is:

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right) = \left\{\begin{array}{l}\mathrm{g}\left(\mathrm{x}\right), \mathrm{if} \mathrm{x}<2\\ \mathrm{h}\left(\mathrm{x}\right), \mathrm{if} \mathrm{x}\ge 2\end{array}\right. \\ \mathrm{and} \mathrm{g}\left(2\right) = \mathrm{h}\left(2\right) \end{gathered}$$

Step 2. Result

$$\begin{gathered} \mathrm{Since} \mathrm{g}(2) = \mathrm{h}(2) \mathrm{it} \mathrm{means} \mathrm{that} \mathrm{the} \mathrm{function} \mathrm{is} \mathrm{continuous} \mathrm{at} \mathrm{x}=2 \mathrm{but} \\ \mathrm{it} \mathrm{is} \mathrm{not} \mathrm{necessary} \mathrm{that} \mathrm{it} \mathrm{is} \mathrm{differentiable} \mathrm{at} \mathrm{x} = 2. \\ \mathrm{For} \mathrm{this}, \mathrm{following} \mathrm{example} \mathrm{can} \mathrm{be} \mathrm{taken}: \\ \mathrm{f}\left(\mathrm{x}\right) = \left\{\begin{array}{l}2\mathrm{x}+3, \mathrm{if} \mathrm{x}<2\\ {\mathrm{x}}^2+3, \mathrm{if} \mathrm{x}\ge 2\end{array}\right. \\ \mathrm{Hence}, \mathrm{the} \mathrm{function} \mathrm{is} \mathrm{continuous} \mathrm{at} \mathrm{x}=2 \mathrm{but} \mathrm{not} \mathrm{differentiable} \mathrm{at} \mathrm{x}=2. \end{gathered}$$