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Differentiability at the right endpoint x = 2

The function f(x)=4−x2f(x) = \sqrt{4 - x^2} has domain [−2,2][-2, 2] and its graph is the upper half of a circle of radius 2 centered at the origin.

Consider f(x)=4−x2f(x) = \sqrt{4 - x^2} on its domain [−2,2][-2,2].

Differentiability at the right endpoint x = 2 graph

Using the limit definition f′(2)=lim⁡x→2f(x)−f(2)x−2f'(2) = \lim_{x \to 2} \dfrac{f(x) - f(2)}{x-2} (with x→2−x \to 2^-, since ff is only defined for x≤2x \le 2), determine whether ff is differentiable at x=2x = 2. State the value of the limit if it exists, or explain why it does not.

Hint

Compute f(2)=0f(2)=0, form the difference quotient 4−x2x−2\dfrac{\sqrt{4-x^2}}{x-2}, factor 4−x2=(2−x)(2+x)4-x^2=(2-x)(2+x), simplify, and then let x→2−x \to 2^-. Watch what happens to the denominator as x→2x\to 2.

Step-by-step solution

Since f(2)=4−4=0f(2) = \sqrt{4-4} = 0, the difference quotient is 4−x2−0x−2=(2−x)(2+x)x−2\dfrac{\sqrt{4-x^2} - 0}{x - 2} = \dfrac{\sqrt{(2-x)(2+x)}}{x-2}. Rewrite x−2=−(2−x)x - 2 = -(2-x), so the quotient becomes 2−x2+x−(2−x)=2+x−2−x\dfrac{\sqrt{2-x}\sqrt{2+x}}{-(2-x)} = \dfrac{\sqrt{2+x}}{-\sqrt{2-x}}.

As x→2−x \to 2^-, 2−x→0+2 - x \to 0^+, so 2−x→0+\sqrt{2-x} \to 0^+ while 2+x→2\sqrt{2+x} \to 2. Thus the quotient 2+x−2−x→−∞\dfrac{\sqrt{2+x}}{-\sqrt{2-x}} \to -\infty, so the limit does not exist as a finite number. This means ff is not differentiable at x=2x=2; graphically the tangent line there is vertical (infinite slope).

Differentiability at the left endpoint x = -2

Now consider the same function f(x)=4−x2f(x) = \sqrt{4 - x^2} at the left endpoint x=−2x = -2. Using the limit definition f′(−2)=lim⁡x→−2f(x)−f(−2)x−(−2)f'(-2) = \lim_{x \to -2} \dfrac{f(x) - f(-2)}{x-(-2)} (with x→−2+x \to -2^+, since ff is only defined for x≥−2x \ge -2), determine whether ff is differentiable at x=−2x = -2, and state whether the limit exists.

Hint

Follow the same simplification as before: use f(−2)=0f(-2) = 0, factor 4−x2=(2−x)(2+x)4 - x^2 = (2-x)(2+x), and simplify 4−x2x+2\dfrac{\sqrt{4-x^2}}{x+2} before taking the one-sided limit x→−2+x \to -2^+.

Step-by-step solution

Since f(−2)=4−4=0f(-2) = \sqrt{4-4} = 0, the difference quotient is 4−x2x+2=(2−x)(2+x)x+2=2−x2+xx+2\dfrac{\sqrt{4-x^2}}{x+2} = \dfrac{\sqrt{(2-x)(2+x)}}{x+2} = \dfrac{\sqrt{2-x}\sqrt{2+x}}{x+2}. For x>−2x > -2 we have x+2>0x+2 > 0, so x+2=(2+x)2x+2 = (\sqrt{2+x})^2. Therefore 2−x2+x(2+x)2=2−x2+x\dfrac{\sqrt{2-x}\sqrt{2+x}}{(\sqrt{2+x})^2} = \dfrac{\sqrt{2-x}}{\sqrt{2+x}}.

As x→−2+x \to -2^+, 2+x→0+2+x \to 0^+ so 2+x→0+\sqrt{2+x}\to 0^+, while 2−x→2\sqrt{2-x} \to 2. Thus the quotient tends to +∞+\infty, so the limit does not exist finitely. Therefore ff is not differentiable at x=−2x=-2 either; the graph has a vertical tangent line at this endpoint as well.

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