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Differentiability at a piecewise junction point

Consider the piecewise function f(x)={x2x≤12x−1x>1f(x) = \begin{cases} x^2 & x \le 1 \\ 2x - 1 & x > 1 \end{cases}

Determine whether f(x)f(x) is differentiable at x=1x = 1. To justify your answer, compute the left-hand derivative lim⁡h→0−f(1+h)−f(1)h\lim_{h \to 0^-} \dfrac{f(1+h) - f(1)}{h} and the right-hand derivative lim⁡h→0+f(1+h)−f(1)h\lim_{h \to 0^+} \dfrac{f(1+h) - f(1)}{h} separately, then state your conclusion.

Differentiability at a piecewise junction point graph
Hint

For h<0h < 0, 1+h≤11+h \le 1 so use f(1+h)=(1+h)2f(1+h) = (1+h)^2. For h>0h > 0, 1+h>11+h > 1 so use f(1+h)=2(1+h)−1f(1+h) = 2(1+h)-1. Compare the two one-sided limits of the difference quotient.

Step-by-step solution

For h→0−h \to 0^-, since 1+h≤11+h \le 1, we use f(1+h)=(1+h)2f(1+h) = (1+h)^2 and f(1)=1f(1) = 1. So the difference quotient is (1+h)2−1h=2h+h2h=2+h\dfrac{(1+h)^2 - 1}{h} = \dfrac{2h + h^2}{h} = 2 + h, which tends to 22 as h→0−h \to 0^-.

For h→0+h \to 0^+, since 1+h>11+h > 1, we use f(1+h)=2(1+h)−1=2h+1f(1+h) = 2(1+h) - 1 = 2h + 1, and f(1)=1f(1) = 1. So the difference quotient is (2h+1)−1h=2hh=2\dfrac{(2h+1) - 1}{h} = \dfrac{2h}{h} = 2, which tends to 22 as h→0+h \to 0^+. Since both one-sided limits equal 22, the limit exists and equals 22, so ff is differentiable at x=1x=1 with f′(1)=2f'(1) = 2.

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