krit.club logo

Calculus - Continuity and differentiability (HL)

Grade 12IB_AA

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

A function f(x)f(x) is continuous at a point x=ax = a if the following three conditions are met: f(a)f(a) is defined, lim⁡x→af(x)\lim_{x \to a} f(x) exists, and lim⁡x→af(x)=f(a)\lim_{x \to a} f(x) = f(a).

•

Continuity on an interval [a,b][a, b] requires the function to be continuous at every point in the open interval (a,b)(a, b) and satisfy one-sided continuity at the endpoints: lim⁡x→a+f(x)=f(a)\lim_{x \to a^+} f(x) = f(a) and lim⁡x→b−f(x)=f(b)\lim_{x \to b^-} f(x) = f(b).

•

A function f(x)f(x) is differentiable at x=ax = a if the limit f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} exists and is finite.

•

Relationship between Continuity and Differentiability: If a function is differentiable at x=ax = a, it must be continuous at x=ax = a. However, continuity does not guarantee differentiability (e.g., f(x)=∣x∣f(x) = |x| is continuous at x=0x=0 but not differentiable there).

•

For a piecewise function to be differentiable at a junction x=kx = k, the left-hand derivative (LHD) must equal the right-hand derivative (RHD), and the function must be continuous at x=kx = k.

•

A function fails to be differentiable at points where there is a 'sharp corner' (cusp), a vertical tangent, or a point of discontinuity.

📐Formulae

lim⁡x→a−f(x)=lim⁡x→a+f(x)=f(a)\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a) (Condition for Continuity)

f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} (Derivative from First Principles)

f′(a)=lim⁡x→af(x)−f(a)x−af'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} (Alternative Derivative Definition)

💡Examples

Problem 1:

Find the values of aa and bb such that the function f(x)={ax2+10x<2x2−6x+bx≥2f(x) = \begin{cases} ax^2 + 10 & x < 2 \\ x^2 - 6x + b & x \geq 2 \end{cases} is differentiable at x=2x = 2.

Solution:

  1. For differentiability, f(x)f(x) must first be continuous at x=2x = 2. Therefore, the limits from both sides must be equal: lim⁡x→2−(ax2+10)=lim⁡x→2+(x2−6x+b)\lim_{x \to 2^-} (ax^2 + 10) = \lim_{x \to 2^+} (x^2 - 6x + b) a(2)2+10=(2)2−6(2)+ba(2)^2 + 10 = (2)^2 - 6(2) + b 4a+10=4−12+b  ⟹  4a+10=−8+b  ⟹  4a−b=−18(Eq.1)4a + 10 = 4 - 12 + b \implies 4a + 10 = -8 + b \implies 4a - b = -18 \quad (Eq. 1)
  2. Now, the derivatives from both sides must be equal at x=2x = 2. For x<2,f′(x)=2axx < 2, f'(x) = 2ax. For x>2,f′(x)=2x−6x > 2, f'(x) = 2x - 6. Equating them at x=2x = 2: 2a(2)=2(2)−6  ⟹  4a=−2  ⟹  a=−122a(2) = 2(2) - 6 \implies 4a = -2 \implies a = -\frac{1}{2}
  3. Substitute a=−12a = -\frac{1}{2} into Eq. 1: 4(−12)−b=−18  ⟹  −2−b=−18  ⟹  b=164(-\frac{1}{2}) - b = -18 \implies -2 - b = -18 \implies b = 16.

Explanation:

To ensure differentiability, we satisfy two conditions: continuity (the pieces meet at the same yy-value) and smoothness (the slopes of the pieces are equal at the junction).

Problem 2:

Show that the function f(x)=∣x−3∣f(x) = |x - 3| is not differentiable at x=3x = 3.

Solution:

We check the limit for the derivative using the piecewise definition: f(x)={x−3x≥3−(x−3)x<3f(x) = \begin{cases} x - 3 & x \geq 3 \\ -(x - 3) & x < 3 \end{cases} Find the Right-Hand Derivative (RHD): f′(3+)=lim⁡h→0+f(3+h)−f(3)h=lim⁡h→0+(3+h−3)−0h=lim⁡h→0+hh=1f'(3^+) = \lim_{h \to 0^+} \frac{f(3+h) - f(3)}{h} = \lim_{h \to 0^+} \frac{(3+h-3) - 0}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1 Find the Left-Hand Derivative (LHD): f′(3−)=lim⁡h→0−f(3+h)−f(3)h=lim⁡h→0−−(3+h−3)−0h=lim⁡h→0−−hh=−1f'(3^-) = \lim_{h \to 0^-} \frac{f(3+h) - f(3)}{h} = \lim_{h \to 0^-} \frac{-(3+h-3) - 0}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1 Since LHD≠RHDLHD \neq RHD, the derivative does not exist at x=3x = 3.

Explanation:

At x=3x=3, the graph of the absolute value function has a sharp corner (vertex). While the function is continuous there, the instantaneous rate of change is different when approaching from the left versus the right.