Differentiability of Piecewise Functions - Part(4)

These calculus questions focus on the differentiability of functions. Each problem is presented with its solution hidden inside a collapsible dropdown so you can practice independently before reviewing the steps[cite: 1].

Theorems

Theorem 1. If a function \( f \) is differentiable at \( x = a \), then \( f \) is continuous at \( x = a \).

Contrapositive. If \( f \) is not continuous at \( x = a \), then \( f \) is not differentiable at \( x = a \).

Theorem 2. If \( f \) is continuous at \( x = a \) and

\[ \lim_{x \to a^+} f'(x) = \lim_{x \to a^-} f'(x), \]

then \( f \) is differentiable at \( x = a \) and

\[ f'(a) = \lim_{x \to a^+} f'(x) = \lim_{x \to a^-} f'(x). \]

Questions and Solutions

Question 1

Determine whether the function \[ f(x) = \begin{cases} 2x^2, & x \le 1 \\ 2\sqrt{x}, & x > 1 \end{cases} \] is differentiable at \( x = 1 \).

View Solution
  • Evaluate continuity at \( x = 1 \): \[ f(1) = 2(1)^2 = 2 \] \[ \lim_{x \to 1^-} f(x) = 2, \quad \lim_{x \to 1^+} f(x) = 2 \]
  • Since the limits equal \( f(1) \), the function is continuous at \( x = 1 \).
  • Compute derivatives: \[ f'(x) = 4x \quad (x < 1), \qquad f'(x) = \frac{1}{\sqrt{x}} \quad (x > 1) \]
  • Evaluate derivative limits: \[ \lim_{x \to 1^-} f'(x) = 4, \quad \lim_{x \to 1^+} f'(x) = 1 \]
  • Since the limits are not equal, \( f'(1) \) does not exist. The function is not differentiable at \( x = 1 \).

Question 2

Let \[ f(x) = \begin{cases} x^3, & x \le 0 \\ x^3 + 1, & x > 0 \end{cases} \] Show that although \[ \lim_{x \to 0^+} f'(x) = \lim_{x \to 0^-} f'(x), \] the derivative \( f'(0) \) does not exist.

View Solution
  • \[ \lim_{x \to 0^-} f(x) = 0, \quad \lim_{x \to 0^+} f(x) = 1 \]
  • The function is not continuous at \( x = 0 \), hence not differentiable there.
  • For all \( x \neq 0 \): \[ f'(x) = 3x^2 \]
  • \[ \lim_{x \to 0^-} f'(x) = \lim_{x \to 0^+} f'(x) = 0 \]
  • Even though the derivative limits agree, \( f'(0) \) does not exist due to discontinuity.

Question 3

Find constants \( A \) and \( B \) such that \[ f(x) = \begin{cases} 2x^2, & x \le 2 \\ Ax + B, & x > 2 \end{cases} \] is differentiable at \( x = 2 \).

View Solution
  • Continuity at \( x = 2 \): \[ \lim_{x \to 2^-} f(x) = 8, \quad \lim_{x \to 2^+} f(x) = 2A + B \]
  • \[ 2A + B = 8 \]
  • Derivatives: \[ f'(x) = 4x \ (x<2), \quad f'(x) = A \ (x>2) \]
  • \[ \lim_{x \to 2^-} f'(x) = 8, \quad \lim_{x \to 2^+} f'(x) = A \]
  • Thus \( A = 8 \) and \( B = -8 \).

Question 4

Find all values of \( x \) for which \[ f(x) = \sqrt{x^2 - 2x + 1} \] is not differentiable.

View Solution
  • \[ f(x) = \sqrt{(x-1)^2} = |x-1| \]
  • \[ f(x) = \begin{cases} x - 1, & x > 1 \\ -(x - 1), & x < 1 \end{cases} \]
  • The function is continuous at \( x = 1 \).
  • Derivatives: \[ f'(x) = 1 \ (x<1), \quad f'(x) = -1 \ (x>1) \]
  • The derivative limits differ, so \( f \) is not differentiable at \( x = 1 \).

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