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Differentiability of |x^2 - 4| at x = 2

Consider the function f(x)=∣x2−4∣f(x) = |x^2 - 4|. Note that near x=2x = 2, x2−4≥0x^2 - 4 \geq 0 for x≥2x \geq 2 and x2−4<0x^2 - 4 < 0 for x<2x < 2 (and similarly, with signs reversed, near x=−2x = -2), so ff is defined piecewise by x2−4x^2-4 and 4−x24-x^2 on either side of these points.

Differentiability of |x^2 - 4| at x = 2 graph

Using the limit definition of the derivative, f′(2)=lim⁡x→2f(x)−f(2)x−2,f'(2) = \lim_{x \to 2} \frac{f(x) - f(2)}{x-2}, determine whether f(x)=∣x2−4∣f(x) = |x^2-4| is differentiable at x=2x = 2. If it is not differentiable, state why (referring to the graph's behavior at that point), and give the values of the left-hand and right-hand limits of the difference quotient.

Hint

Near x=2x=2, write f(x)=4−x2f(x) = 4 - x^2 for xx slightly less than 22 and f(x)=x2−4f(x) = x^2 - 4 for xx slightly greater than 22. Compute the one-sided limits of f(x)−f(2)x−2\frac{f(x)-f(2)}{x-2} separately and compare them.

Step-by-step solution

For xx near 22 but less than 22, x2−4<0x^2 - 4 < 0 so f(x)=4−x2f(x) = 4 - x^2, giving f(x)−f(2)x−2=4−x2−0x−2=−(x2−4)x−2=−(x−2)(x+2)x−2=−(x+2)→−4\frac{f(x)-f(2)}{x-2} = \frac{4-x^2 - 0}{x-2} = \frac{-(x^2-4)}{x-2} = \frac{-(x-2)(x+2)}{x-2} = -(x+2) \to -4 as x→2−x \to 2^-. For xx near 22 but greater than 22, x2−4≥0x^2-4 \geq 0 so f(x)=x2−4f(x) = x^2-4, giving f(x)−f(2)x−2=x2−4−0x−2=(x−2)(x+2)x−2=x+2→4\frac{f(x)-f(2)}{x-2} = \frac{x^2-4-0}{x-2} = \frac{(x-2)(x+2)}{x-2} = x+2 \to 4 as x→2+x \to 2^+.

Since the left-hand limit (−4-4) does not equal the right-hand limit (44), the two-sided limit defining f′(2)f'(2) does not exist. Thus ff is not differentiable at x=2x=2; graphically this corresponds to the sharp corner visible in the plot at x=2x=2, where the tangent line slope jumps abruptly rather than varying continuously.

Differentiability of |x^2 - 4| at x = -2

Using the same idea as before, determine whether f(x)=∣x2−4∣f(x) = |x^2-4| is differentiable at x=−2x = -2 by examining lim⁡x→−2f(x)−f(−2)x−(−2).\lim_{x \to -2} \frac{f(x) - f(-2)}{x-(-2)}. State whether the function is differentiable there, and if not, give the left-hand and right-hand limit values of the difference quotient.

Hint

Near x=−2x=-2, determine on which side x2−4x^2-4 is positive versus negative, write f(x)f(x) accordingly, and compute the one-sided limits of the difference quotient separately.

Step-by-step solution

For xx slightly less than −2-2 (so x<−2x < -2), x2−4>0x^2-4>0, so f(x)=x2−4f(x) = x^2-4, and f(x)−f(−2)x+2=x2−4−0x+2=(x−2)(x+2)x+2=x−2→−4\frac{f(x)-f(-2)}{x+2} = \frac{x^2-4-0}{x+2} = \frac{(x-2)(x+2)}{x+2} = x-2 \to -4 as x→−2−x\to -2^-. For xx slightly greater than −2-2 (so −2<x<2-2 < x < 2), x2−4<0x^2-4 < 0, so f(x)=4−x2f(x)=4-x^2, and f(x)−f(−2)x+2=4−x2−0x+2=−(x2−4)x+2=−(x−2)(x+2)x+2=−(x−2)→4\frac{f(x)-f(-2)}{x+2} = \frac{4-x^2-0}{x+2} = \frac{-(x^2-4)}{x+2} = \frac{-(x-2)(x+2)}{x+2} = -(x-2) \to 4 as x→−2+x \to -2^+.

The left-hand limit is −4-4 and the right-hand limit is 44; since these differ, f′(−2)f'(-2) does not exist. So ff is not differentiable at x=−2x = -2 either, consistent with the corner seen in the graph at that point.

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