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Module II — Continuity and Differentiation

Module II — Continuity and Differentiation

Course: MAT1CJ101 — Differential Calculus
Hours: 15 | Textbook Sections: §1.5, §2.1, §2.2, §2.3, §2.5, §2.6 (Thomas & Finney, 9th Ed.)

1. Continuity

1.1 Definition

A function $f$ is continuous at a point $x = c$ if three conditions hold simultaneously:

  1. $f(c)$ is defined (the function has a value at $c$)
  2. $\lim_{x \to c} f(x)$ exists (the limit exists)
  3. $\lim_{x \to c} f(x) = f(c)$ (the limit equals the value)

If any condition fails, $f$ has a discontinuity at $c$.

Intuition: Continuity means you can draw the graph through $(c, f(c))$ without lifting your pencil. A discontinuity is a break, jump, or hole.

1.2 Types of Discontinuity

Type What happens Example
Removable Limit exists but $\neq f(c)$, or $f(c)$ undefined $f(x) = \dfrac{x^2-4}{x-2}$ at $x = 2$
Jump Left and right limits exist but differ $f(x) = \dfrac{|x|}{x}$ at $x = 0$
Infinite $f(x) \to \pm\infty$ as $x \to c$ $f(x) = \dfrac{1}{x}$ at $x = 0$
Oscillating Limit does not exist (not a jump) $f(x) = \sin(1/x)$ at $x = 0$

Removable discontinuities can be “fixed” by redefining $f(c) = L = \lim_{x \to c} f(x)$. This is called a continuous extension.

Worked Example 1: Determine whether $f(x) = \dfrac{x^2 + x - 6}{x^2 - 4}$ is continuous at $x = 2$.

Factor: $x^2 + x - 6 = (x+3)(x-2)$, and $x^2 - 4 = (x+2)(x-2)$.

\[f(x) = \frac{(x+3)(x-2)}{(x+2)(x-2)} = \frac{x+3}{x+2} \quad \text{for } x \neq 2.\]

$f(2)$ is undefined ($0/0$). But $\displaystyle\lim_{x \to 2} \frac{x+3}{x+2} = \frac{5}{4}$.

Discontinuity at $x = 2$, but it is removable: define $f(2) = 5/4$. The extended function is continuous at $x = 2$. (Oct 2024, Q19b)

1.3 Continuity on an Interval

  • $f$ is continuous on an open interval $(a, b)$ if it is continuous at every interior point.
  • $f$ is continuous on $[a, b]$ if it is continuous on $(a, b)$, and:
    • $\lim_{x \to a^+} f(x) = f(a)$
    • $\lim_{x \to b^-} f(x) = f(b)$

Theorem: Polynomials are continuous everywhere. Rational functions are continuous wherever their denominator is non-zero.

1.4 The Intermediate Value Theorem (IVT)

Theorem (IVT): If $f$ is continuous on $[a, b]$ and $v$ is any value between $f(a)$ and $f(b)$, then there exists at least one $c$ in $(a, b)$ with $f(c) = v$.

Why this is useful: To show a zero exists, show $f$ changes sign on an interval. To show an equation has a solution, rearrange it into the form $f(x) = 0$ and apply IVT.

Worked Example 2: Is any real number exactly 1 less than its cube? (Oct 2024, Q12)

We want to show $x^3 - x - 1 = 0$ has a solution.

Let $f(x) = x^3 - x - 1$. $f$ is a polynomial, so continuous everywhere.

\(f(1) = 1 - 1 - 1 = -1 < 0\) \(f(2) = 8 - 2 - 1 = 5 > 0\)

By IVT, there exists $c \in (1, 2)$ with $f(c) = 0$. So yes, such a real number exists (it is between 1 and 2).

2. The Derivative of a Function

2.1 Motivation — Instantaneous Rate of Change

The average rate of change of $f$ over $[x_0, x_1]$ is $\dfrac{\Delta y}{\Delta x} = \dfrac{f(x_1) - f(x_0)}{x_1 - x_0}$.

As $x_1 \to x_0$, this ratio approaches the instantaneous rate of change at $x_0$ — the derivative.

2.2 Definition of the Derivative

The derivative of $f$ at $x_0$ is:

\[f'(x_0) = \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h}\]

provided this limit exists. If it exists, $f$ is differentiable at $x_0$.

Alternate form (with $x \to x_0$):

\[f'(x_0) = \lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0}\]

Geometric meaning: $f’(x_0)$ is the slope of the tangent line to $y = f(x)$ at $(x_0, f(x_0))$.

Equation of tangent line at $(x_0, f(x_0))$:

\[y - f(x_0) = f'(x_0)(x - x_0)\]

Equation of normal line (perpendicular to tangent):

\[y - f(x_0) = -\frac{1}{f'(x_0)}(x - x_0) \quad [\text{when } f'(x_0) \neq 0]\]

2.3 Computing Derivatives from the Definition

Worked Example 3: Find $f’(x)$ for $f(x) = x^2$.

\[f'(x) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h}\] \[= \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h}\] \[= \lim_{h \to 0} \frac{2xh + h^2}{h}\] \[= \lim_{h \to 0} (2x + h)\] \[= 2x\]

Worked Example 4: Find $f’(x)$ for $f(x) = 1/x$ ($x \neq 0$).

\[f'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h}\] \[= \lim_{h \to 0} \frac{x - (x+h)}{x(x+h)h}\] \[= \lim_{h \to 0} \frac{-h}{x(x+h)h}\] \[= \lim_{h \to 0} \frac{-1}{x(x+h)}\] \[= -\frac{1}{x^2}\]

2.4 Differentiability and Continuity

Theorem: If $f$ is differentiable at $c$, then $f$ is continuous at $c$.

Contrapositive: If $f$ is not continuous at $c$, it cannot be differentiable there.

The converse is false: Continuity does not imply differentiability. Example: $f(x) = x $ is continuous at $0$ but not differentiable there (the graph has a corner).

Where can differentiability fail?

  • Corners (like $ x $ at $0$)
  • Cusps (like $x^{2/3}$ at $0$)
  • Vertical tangents (slope $\to \pm\infty$)
  • Jump discontinuities

3. Differentiation Rules

3.1 Basic Rules

Let $f$ and $g$ be differentiable, $k$ a constant.

Rule Formula
Constant $(k)’ = 0$
Power $(x^n)’ = nx^{n-1}$
Constant multiple $(kf)’ = kf’$
Sum $(f + g)’ = f’ + g’$
Difference $(f - g)’ = f’ - g’$

Power Rule works for any real exponent $n$ (proof for integers from the definition; for rationals/reals, proved later).

Worked Example 5: Differentiate $y = 4x^3 - 7x^2 + 2x - 9$.

\[\frac{dy}{dx} = 4(3x^2) - 7(2x) + 2(1) - 0 = 12x^2 - 14x + 2\]

3.2 Product Rule

Theorem: $(fg)’ = f’g + fg’$

Intuition: Both factors change simultaneously. The rate of change of their product accounts for each factor’s change while the other is momentarily fixed.

Worked Example 6: Differentiate $y = (x^2 + 1)(3x - 5)$.

Let $f = x^2 + 1$, $g = 3x - 5$. $f’ = 2x$, $g’ = 3$.

\[y' = f'g + fg' = 2x(3x - 5) + (x^2 + 1)(3) = 6x^2 - 10x + 3x^2 + 3 = 9x^2 - 10x + 3\]

(Verify by expanding first: $y = 3x^3 - 5x^2 + 3x - 5$, then $y’ = 9x^2 - 10x + 3$. ✓)

3.3 Quotient Rule

Theorem: $(f/g)’ = \dfrac{f’g - fg’}{g^2}$

Mnemonic: “Low dee-high minus high dee-low, square the bottom and away we go.”

Worked Example 7: Differentiate $y = \dfrac{x^2 - 1}{x + 1}$.

$f = x^2 - 1$, $g = x + 1$. $f’ = 2x$, $g’ = 1$.

\[y' = \frac{2x(x+1) - (x^2-1)(1)}{(x+1)^2} = \frac{2x^2 + 2x - x^2 + 1}{(x+1)^2} = \frac{x^2 + 2x + 1}{(x+1)^2} = \frac{(x+1)^2}{(x+1)^2} = 1 \quad (\text{for } x \neq -1)\]

(Makes sense: $y = \dfrac{(x+1)(x-1)}{x+1} = x - 1$ for $x \neq -1$, so $y’ = 1$. ✓)

3.4 Summary of Rules so Far

Rule Formula
Power $\dfrac{d}{dx}[x^n] = nx^{n-1}$
Sum/Difference $(f \pm g)’ = f’ \pm g’$
Constant multiple $(cf)’ = cf’$
Product $(fg)’ = f’g + fg’$
Quotient $(f/g)’ = \dfrac{f’g - fg’}{g^2}$

4. Rates of Change

The derivative as a rate of change interprets $f’(x)$ as the instantaneous rate at which $f$ changes per unit change in $x$.

  • Position → Velocity: If $s(t)$ is position at time $t$, then $v(t) = s’(t)$ is instantaneous velocity.
  • Velocity → Acceleration: $a(t) = v’(t) = s’‘(t)$.
  • Marginal cost: If $C(q)$ is cost to produce $q$ units, $C’(q)$ is marginal cost (cost of one more unit).

Worked Example 8: Find the average rate of change of $f(t) = 1/t$ from $t = 2$ to $t = 3$. (Oct 2024, Q14)

\[\text{Average rate} = \frac{f(3) - f(2)}{3 - 2} = \frac{\frac{1}{3} - \frac{1}{2}}{1} = -\frac{1}{6}\]

Worked Example 9: A ball is dropped. Its height is $h(t) = 100 - 5t^2$ metres after $t$ seconds. Find its velocity at $t = 3$ s.

\[h'(t) = -10t\] \[h'(3) = -30 \text{ m/s} \quad (\text{negative = falling downward})\]

4.1 Horizontal Tangents

A curve $y = f(x)$ has a horizontal tangent at points where $f’(x) = 0$.

Worked Example 10: Does $y = x^4 - 2x^2 + 2$ have horizontal tangents? (Oct 2024, Q3)

\[y' = 4x^3 - 4x = 4x(x^2 - 1) = 4x(x-1)(x+1)\]

$y’ = 0$ at $x = 0$, $x = 1$, $x = -1$.

Yes, three horizontal tangents:

  • At $x = 0$: $y = 2$ → point $(0, 2)$
  • At $x = 1$: $y = 1$ → point $(1, 1)$
  • At $x = -1$: $y = 1$ → point $(-1, 1)$

5. The Chain Rule

5.1 Statement

If $y = f(g(x))$, then:

\[\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\]

Or in Leibniz notation, if $y = f(u)$ and $u = g(x)$:

\[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\]

Intuition: If $u$ changes at rate $g’(x)$ per unit $x$, and $y$ changes at rate $f’(u)$ per unit $u$, then $y$ changes at rate $f’(u) \cdot g’(x)$ per unit $x$.

5.2 Applying the Chain Rule

Identify the outer function and inner function, then differentiate outer (leaving inner alone) times derivative of inner.

Worked Example 11: Differentiate $y = (x^2 + 1)^5$.

Outer: $u^5$ where $u = x^2 + 1$. Outer derivative: $5u^4$. Inner derivative: $2x$.

\[y' = 5(x^2+1)^4 \cdot 2x = 10x(x^2+1)^4\]

Worked Example 12: Differentiate $y = \sqrt{3x^2 - 1}$.

\[y = (3x^2 - 1)^{1/2}\]

Outer: $u^{1/2}$, derivative $= \frac{1}{2}u^{-1/2}$. Inner: $3x^2 - 1$, derivative $= 6x$.

\[y' = \frac{1}{2}(3x^2-1)^{-1/2} \cdot 6x = \frac{3x}{\sqrt{3x^2-1}}\]

Worked Example 13: Differentiate $y = \dfrac{(2x + 1)^3}{(x - 1)^2}$ using chain rule and quotient rule.

Numerator: $u = (2x+1)^3$, $u’ = 3(2x+1)^2 \cdot 2 = 6(2x+1)^2$. Denominator: $v = (x-1)^2$, $v’ = 2(x-1)$.

Quotient rule: $y’ = \dfrac{u’v - uv’}{v^2}$

\[= \frac{6(2x+1)^2(x-1)^2 - (2x+1)^3 \cdot 2(x-1)}{(x-1)^4}\]

Factor out $(2x+1)^2(x-1)$:

\[= \frac{(2x+1)^2(x-1)[6(x-1) - 2(2x+1)]}{(x-1)^4} = \frac{(2x+1)^2[6x - 6 - 4x - 2]}{(x-1)^3} = \frac{(2x+1)^2(2x - 8)}{(x-1)^3} = \frac{2(2x+1)^2(x - 4)}{(x-1)^3}\]

5.3 Repeated Chain Rule

For compositions of more than two functions: $y = f(g(h(x)))$:

\[y' = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)\]

Worked Example 14: Differentiate $y = \sqrt{1 + \sqrt{x}}$.

\[y = (1 + x^{1/2})^{1/2}\]

Outer: $u^{1/2}$, derivative $= \frac{1}{2}u^{-1/2}$. Middle: $1 + x^{1/2}$, derivative $= \frac{1}{2}x^{-1/2}$.

\[y' = \frac{1}{2}(1+\sqrt{x})^{-1/2} \cdot \frac{1}{2}x^{-1/2} = \frac{1}{4\sqrt{x} \cdot \sqrt{1+\sqrt{x}}}\]

6. Implicit Differentiation and Rational Exponents

6.1 Implicit vs. Explicit Functions

An explicit function is given directly as $y = f(x)$. An implicit function is defined by an equation $F(x, y) = 0$ where $y$ is not isolated.

Example: $x^2 + y^2 = 25$ defines a circle. We cannot write $y = f(x)$ for the whole circle, but at each point we can find $\dfrac{dy}{dx}$.

6.2 The Method of Implicit Differentiation

Steps:

  1. Differentiate both sides of $F(x, y) = 0$ with respect to $x$.
  2. Treat $y$ as a function of $x$ — apply the chain rule whenever $y$ appears: $\dfrac{d}{dx}[y^n] = ny^{n-1} \cdot \dfrac{dy}{dx}$.
  3. Solve the resulting equation for $\dfrac{dy}{dx}$.

Worked Example 15: Find $\dfrac{dy}{dx}$ if $2y = x^2 + \sin y$. (Oct 2024, Q5)

Differentiate both sides with respect to $x$:

\[2\frac{dy}{dx} = 2x + \cos(y) \cdot \frac{dy}{dx}\]

Gather $\dfrac{dy}{dx}$ terms:

\[2\frac{dy}{dx} - \cos(y)\frac{dy}{dx} = 2x\] \[\frac{dy}{dx}(2 - \cos y) = 2x\] \[\frac{dy}{dx} = \frac{2x}{2 - \cos y}\]

Worked Example 16: Find the tangent and normal to $x^2 - xy + y^2 = 7$ at $(-1, 2)$. (Oct 2024, Q6)

Differentiate implicitly:

\[2x - \left(y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} = 0\] \[2x - y - x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0\] \[\frac{dy}{dx}(2y - x) = y - 2x\] \[\frac{dy}{dx} = \frac{y - 2x}{2y - x}\]

At $(-1, 2)$:

\[\frac{dy}{dx} = \frac{2 - 2(-1)}{2(2) - (-1)} = \frac{2 + 2}{4 + 1} = \frac{4}{5}\]

Tangent line at $(-1, 2)$: $y - 2 = \dfrac{4}{5}(x + 1) \implies 4x - 5y + 14 = 0$

Normal line (slope $= -\dfrac{5}{4}$): $y - 2 = -\dfrac{5}{4}(x + 1) \implies 5x + 4y - 3 = 0$

6.3 Finding Critical Points Using Implicit Differentiation

A curve has a horizontal tangent where $\dfrac{dy}{dx} = 0$ (numerator = 0, denominator $\neq 0$), and a vertical tangent where $\dfrac{dy}{dx}$ is undefined (denominator = 0, numerator $\neq 0$).

Worked Example 17: Find points on $x^2 + 2y^2 = 1$ with horizontal tangents.

Differentiate: $2x + 4y\dfrac{dy}{dx} = 0 \implies \dfrac{dy}{dx} = -\dfrac{x}{2y}$

Horizontal tangent: $\dfrac{dy}{dx} = 0 \implies x = 0$

Substitute $x = 0$: $2y^2 = 1 \implies y = \pm\dfrac{1}{\sqrt{2}}$

Points: $\left(0, \dfrac{1}{\sqrt{2}}\right)$ and $\left(0, -\dfrac{1}{\sqrt{2}}\right)$

6.4 Rational Exponents and the Power Rule

For any rational exponent $m/n$ (in lowest terms, $n > 0$):

\[\frac{d}{dx}\left[x^{m/n}\right] = \frac{m}{n} x^{m/n - 1}\]

This follows from implicit differentiation: let $y = x^{m/n}$, so $y^n = x^m$. Differentiate both sides.

Worked Example 18: Differentiate $y = x^{3/4}$.

\[\frac{dy}{dx} = \frac{3}{4} x^{3/4 - 1} = \frac{3}{4} x^{-1/4} = \frac{3}{4x^{1/4}}\]

Worked Example 19: Differentiate $y = x^{2/3}$.

\[\frac{dy}{dx} = \frac{2}{3} x^{-1/3} = \frac{2}{3x^{1/3}} = \frac{2}{3\sqrt[3]{x}}\]

Note: this is undefined at $x = 0$ (vertical tangent — the graph has a cusp there).

7. Summary

Concept Key Formula / Rule
Continuity at $c$ $f(c)$ defined, limit exists, limit $= f(c)$
Removable discontinuity Limit exists but $f(c)$ undefined or wrong
IVT $f$ continuous on $[a,b]$, $f(a)$ and $f(b)$ have opposite signs → zero in $(a,b)$
Derivative definition $f’(x) = \lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$
Power rule $(x^n)’ = nx^{n-1}$
Product rule $(fg)’ = f’g + fg’$
Quotient rule $(f/g)’ = \dfrac{f’g - fg’}{g^2}$
Chain rule $(f\circ g)’(x) = f’(g(x)) \cdot g’(x)$
Implicit diff Differentiate both sides w.r.t. $x$, chain rule on $y$
Rational exponent $(x^{m/n})’ = \dfrac{m}{n}x^{m/n-1}$

8. Practice Problems

Sec A style (3 marks each):

  1. Show that $f(x) = x^3 - x^2 + x - 1$ has a zero in the interval $[0, 2]$.

  2. At what points is $f(x) = x - 2 $ not differentiable? Justify.

  3. Using the definition of the derivative, find $f’(x)$ for $f(x) = 3x^2 + 1$.

  4. Does $y = x^3 + x^2 - x + 1$ have any horizontal tangents? If so, where? (Sec A style, from Q3)

  5. Find $\dfrac{dy}{dx}$ if $2y = x^2 + \sin y$. (Oct 2024, Q5)

  6. Find the normal to $x^2 - xy + y^2 = 7$ at $(-1, 2)$. (Oct 2024, Q6)

  7. The curve $y = ax^2 + bx + c$ passes through $(1, 2)$ and is tangent to $y = x$ at the origin. Find $a$, $b$, $c$. (Oct 2024, Q4)

Sec B style (6 marks each):

  1. Show that $f(x) = x^4 + 3x + 1$ has exactly one zero in $[-2, -1]$. (Oct 2024, Q16) (Hint: IVT for existence; show $f$ is strictly increasing using $f’$ to get uniqueness.)

  2. What is implicit differentiation? When do you need it? Give two worked examples with different curve types. (Oct 2024, Q15)

  3. Differentiate: (a) $y = (x^3 + 2x)^4$, (b) $y = \sqrt{x^2 + 1}$, (c) $y = (x+1)^{2/3}$. Show all chain rule steps.

Sec C style (10 marks):

  1. Show that $f(x) = \dfrac{x^2 + x - 6}{x^2 - 4}$ has a continuous extension to $x = 2$, and find that extension. Then determine all other points of discontinuity and classify each. (Oct 2024, Q19b)