krit.club logo

Calculus - Continuity and Differentiability

Grade 12ICSE

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

🔑Concepts

Continuity at a Point: A function f(x)f(x) is continuous at x=cx = c if the limit of the function as xx approaches cc equals the value of the function at cc. Visually, this means there is no break, hole, or jump in the graph at that point; you can draw the curve through x=cx = c without lifting your pencil from the paper.

Types of Discontinuity: Discontinuities are visual interruptions in a graph. A 'Removable Discontinuity' appears as a tiny hole on a smooth line. A 'Jump Discontinuity' occurs when the graph breaks and restarts at a different height (common in piecewise functions). An 'Infinite Discontinuity' occurs when the function approaches a vertical asymptote, shooting off towards \infty or -\infty.

Continuity on an Interval: A function is continuous on a closed interval [a,b][a, b] if it is continuous at every point in the open interval (a,b)(a, b), and the one-sided limits at the endpoints match the function values: limxa+f(x)=f(a)\lim_{x \to a^+} f(x) = f(a) and limxbf(x)=f(b)\lim_{x \to b^-} f(x) = f(b).

Differentiability: A function is differentiable at x=cx = c if the derivative exists at that point, meaning the left-hand derivative equals the right-hand derivative. Visually, a differentiable graph is 'smooth' and has a unique tangent line. If a graph has a 'sharp corner' or 'cusp' (like the point of a 'V'), it is not differentiable there because no single tangent can be defined.

Relationship between Continuity and Differentiability: Every function that is differentiable at a point is necessarily continuous at that point. However, the reverse is not always true. Visually, a continuous curve might have a sharp point (like y=xy = |x| at x=0x=0) where it is not differentiable, even though the line is never broken.

Intermediate Value Theorem (IVT): If a function ff is continuous on [a,b][a, b], it must take every value between f(a)f(a) and f(b)f(b) at least once. Visually, a continuous path connecting two points at different heights must cross every horizontal line between those two heights.

Rolle's Theorem and Mean Value Theorem (MVT): Rolle's Theorem states that if a function is continuous on [a,b][a, b], differentiable on (a,b)(a, b), and f(a)=f(b)f(a) = f(b), there is a point where the tangent is horizontal (slope = 0). Lagrange's MVT generalizes this, stating there is a point cc where the instantaneous slope (tangent) equals the average slope (secant) of the interval. Visually, at some point cc, the tangent line is parallel to the chord joining the endpoints (a,f(a))(a, f(a)) and (b,f(b))(b, f(b)).

📐Formulae

Continuity Condition: limxcf(x)=limxc+f(x)=f(c)\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)

General Derivative: f(x)=limh0f(x+h)f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

Left Hand Derivative (LHD): Lf(c)=limh0f(c+h)f(c)hLf'(c) = \lim_{h \to 0^-} \frac{f(c+h) - f(c)}{h}

Right Hand Derivative (RHD): Rf(c)=limh0+f(c+h)f(c)hRf'(c) = \lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h}

Mean Value Theorem: f(c)=f(b)f(a)baf'(c) = \frac{f(b) - f(a)}{b - a} for some c(a,b)c \in (a, b)

Chain Rule: dydx=dydududx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

💡Examples

Problem 1:

Determine the value of the constant kk so that the function f(x)f(x) is continuous at x=3x = 3: f(x)={x29x3if x3kif x=3f(x) = \begin{cases} \frac{x^2 - 9}{x - 3} & \text{if } x \neq 3 \\ k & \text{if } x = 3 \end{cases}

Solution:

  1. Find the limit of f(x)f(x) as x3x \to 3: limx3x29x3\lim_{x \to 3} \frac{x^2 - 9}{x - 3}. \n2. Simplify the expression: x29x3=(x3)(x+3)x3=x+3\frac{x^2 - 9}{x - 3} = \frac{(x - 3)(x + 3)}{x - 3} = x + 3 (for x3x \neq 3). \n3. Calculate the limit: limx3(x+3)=3+3=6\lim_{x \to 3} (x + 3) = 3 + 3 = 6. \n4. For continuity at x=3x = 3, the limit must equal the function value: limx3f(x)=f(3)\lim_{x \to 3} f(x) = f(3). \n5. Therefore, 6=k6 = k.

Explanation:

To make the function continuous, we fill the 'removable discontinuity' (the hole in the graph) by setting the function value kk equal to the limit of the function as it approaches that point.

Problem 2:

Examine the differentiability of f(x)=x2/3f(x) = x^{2/3} at x=0x = 0.

Solution:

  1. Use the definition of the derivative at x=0x = 0: f(0)=limh0f(0+h)f(0)hf'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h}. \n2. Substitute the function: f(0)=limh0h2/302/3h=limh0h2/3hf'(0) = \lim_{h \to 0} \frac{h^{2/3} - 0^{2/3}}{h} = \lim_{h \to 0} \frac{h^{2/3}}{h}. \n3. Simplify the exponent: h2/3/h=h2/31=h1/3=1h1/3h^{2/3} / h = h^{2/3 - 1} = h^{-1/3} = \frac{1}{h^{1/3}}. \n4. Evaluate the limit: As h0h \to 0, 1h1/3\frac{1}{h^{1/3}} \to \infty. \n5. Since the limit is not a finite real number, f(x)f(x) is not differentiable at x=0x = 0.

Explanation:

Visually, the graph of x2/3x^{2/3} has a 'cusp' (a very sharp point) at the origin. Because the slope becomes infinite as we approach zero, there is no defined tangent line, making it non-differentiable.