Unit 2.4 – Connecting Differentiability and Continuity

Formal Proof (AP® Level)

Given: \(f'(a)\) exists

To Prove: \(f\) is continuous at \(x = a\), i.e., \(\lim_{x \to a} f(x) = f(a)\)

Proof:

1. We need to show: \(\lim_{x \to a} [f(x) - f(a)] = 0\)
2. Multiply and divide by \((x - a)\):
   \[ \lim_{x \to a} [f(x) - f(a)] = \lim_{x \to a} \left[\frac{f(x) - f(a)}{x - a} \cdot (x - a)\right] \]
3. Use limit properties to separate:
   \[ = \left(\lim_{x \to a} \frac{f(x) - f(a)}{x - a}\right) \cdot \left(\lim_{x \to a} (x - a)\right) \]
4. Recognize the derivative:
   \[ = f'(a) \cdot 0 = 0 \quad \checkmark \]

Therefore: \(\lim_{x \to a} f(x) = f(a)\), so \(f\) is continuous at \(x = a\) ∎

Problem: Determine if \(f(x) = |x - 3|\) is differentiable at \(x = 3\)

Solution:

1. Check continuity:
   • \(f(3) = 0\)
   • \(\lim_{x \to 3} |x - 3| = 0\)
   • Continuous at \(x = 3\) ✓
2. Rewrite as piecewise:
   \[ f(x) = \begin{cases} -(x - 3) = 3 - x & x < 3 \\ x - 3 & x \geq 3 \end{cases} \]
3. Find left-hand derivative:
   • For \(x < 3\): \(f(x) = 3 - x\), so \(f'(x) = -1\)
   • \(f'(3^-) = -1\)
4. Find right-hand derivative:
   • For \(x > 3\): \(f(x) = x - 3\), so \(f'(x) = 1\)
   • \(f'(3^+) = 1\)
5. Compare: \(f'(3^-) = -1 \neq 1 = f'(3^+)\)

Conclusion: NOT differentiable at \(x = 3\) because the one-sided derivatives don't match. This is a corner.

Problem: For what value of \(k\) is \(f(x) = \begin{cases} x^2 & x \leq 2 \\ kx & x > 2 \end{cases}\) differentiable at \(x = 2\)?

Solution:

1. Check continuity first:
   • \(\lim_{x \to 2^-} x^2 = 4\)
   • \(\lim_{x \to 2^+} kx = 2k\)
   • For continuity: \(4 = 2k \Rightarrow k = 2\)
2. Check differentiability with \(k = 2\):
   • \(f'(2^-) = 2x\big|_{x=2} = 4\)
   • \(f'(2^+) = 2\)
   • \(4 \neq 2\), so NOT differentiable!
3. Actually, we need BOTH conditions:
   • Continuity: \(2k = 4 \Rightarrow k = 2\)
   • Differentiability: \(4 = k\)
   • These conflict!

Answer: No value of k makes \(f\) differentiable at \(x = 2\) because continuity requires \(k = 2\) but differentiability requires \(k = 4\). This is impossible!

Problem: Find \(a\) and \(b\) so that \(f(x) = \begin{cases} x^2 & x \leq 1 \\ ax + b & x > 1 \end{cases}\) is differentiable at \(x = 1\)

Solution:

1. Continuity condition:
   • \(\lim_{x \to 1^-} x^2 = 1\)
   • \(\lim_{x \to 1^+} (ax + b) = a + b\)
   • Equation 1: \(a + b = 1\)
2. Differentiability condition:
   • \(f'(1^-) = 2x\big|_{x=1} = 2\)
   • \(f'(1^+) = a\)
   • Equation 2: \(a = 2\)
3. Solve the system:
   • From Equation 2: \(a = 2\)
   • Substitute into Equation 1: \(2 + b = 1 \Rightarrow b = -1\)

Answer: \(a = 2\) and \(b = -1\)

Result: \(f(x) = \begin{cases} x^2 & x \leq 1 \\ 2x - 1 & x > 1 \end{cases}\) is differentiable everywhere

Problem: Identify all points where \(f(x) = \sqrt[3]{(x - 2)^2}\) is not differentiable

Solution:

1. Check continuity: \(f(x)\) is continuous everywhere (cube root is defined for all real numbers)
2. Find the derivative:
   \[ f'(x) = \frac{2}{3}(x - 2)^{-1/3} = \frac{2}{3\sqrt[3]{x - 2}} \]
3. Check where \(f'(x)\) doesn't exist:
   • When \(x - 2 = 0\), i.e., \(x = 2\)
   • At \(x = 2\): \(f'(2)\) is undefined (division by zero)
   • As \(x \to 2\), \(|f'(x)| \to \infty\)

Answer: Not differentiable at \(x = 2\)

Reason: This is a cusp (the function has a sharp point where slopes from both sides go to ±∞)

Problem: Show that \(f(x) = x^{1/3}\) has a vertical tangent at \(x = 0\)

Solution:

1. Check continuity:
   • \(\lim_{x \to 0} x^{1/3} = 0 = f(0)\)
   • Continuous at \(x = 0\) ✓
2. Find the derivative:
   \[ f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3x^{2/3}} = \frac{1}{3\sqrt[3]{x^2}} \]
3. Check behavior at \(x = 0\):
   • \(\lim_{x \to 0^+} f'(x) = \lim_{x \to 0^+} \frac{1}{3\sqrt[3]{x^2}} = +\infty\)
   • \(\lim_{x \to 0^-} f'(x) = \lim_{x \to 0^-} \frac{1}{3\sqrt[3]{x^2}} = +\infty\)
   • The slopes blow up to \(+\infty\) from both sides

Conclusion: The function is continuous at \(x = 0\) but not differentiable because the tangent line is vertical (infinite slope)

Determine if the following functions are differentiable at the given point:

1. \(f(x) = |x + 1|\) at \(x = -1\)
2. \(g(x) = x^{2/5}\) at \(x = 0\)
3. \(h(x) = \begin{cases} x^2 + 1 & x < 1 \\ 3x - 1 & x \geq 1 \end{cases}\) at \(x = 1\)
4. \(k(x) = \sqrt{x}\) at \(x = 0\)

Answers:

1. NOT differentiable — Corner at \(x = -1\) (left derivative = -1, right derivative = +1)
2. NOT differentiable — Cusp at \(x = 0\) (derivative: \(f'(x) = \frac{2}{5}x^{-3/5} \to \pm\infty\) as \(x \to 0\))
3. NOT differentiable — Continuous (both limits = 2), but \(h'(1^-) = 2\) and \(h'(1^+) = 3\) (don't match)
4. NOT differentiable — Vertical tangent at \(x = 0\) from the right (\(k'(x) = \frac{1}{2\sqrt{x}} \to \infty\) as \(x \to 0^+\))