M1: Algebra and Calculus | DSE

Solving linear inequalities follows the same principles as solving linear equations, with one critical difference: multiplying or dividing both sides by a negative number reverses the inequality sign.

$3x - 7 < 2x + 5$ $x < 12$

To solve $ax^2 + bx + c > 0$ (or $< 0$, $\geq 0$, $\leq 0$):

1. Find the roots of the equation $ax^2 + bx + c = 0$
2. Determine the sign of the quadratic in each region defined by the roots
3. Select the regions satisfying the inequality

Example: Solve $x^2 - 5x + 6 \leq 0$.

Factorising: $(x-2)(x-3) \leq 0$. The parabola opens upward (positive $x^2$ coefficient). The expression is non-positive between the roots:

$2 \leq x \leq 3$

For inequalities involving rational expressions, bring all terms to one side and combine into a single fraction. Find the critical values (zeros and undefined points), then test each region.

Example: Solve $\frac{x-1}{x+2} > 0$.

Critical values: $x = 1$ and $x = -2$.

Testing intervals:

• $x < -2$: Both numerator and denominator negative — ratio is positive
• $-2 < x < 1$: Numerator negative, denominator positive — ratio is negative
• $x > 1$: Both positive — ratio is positive

Solution: $x < -2$ or $x > 1$, i.e., $x \in (-\infty, -2) \cup (1, \infty)$.

$|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}$

$|ab| = |a| \cdot |b|$ $|a + b| \leq |a| + |b| \quad \text{(triangle inequality)}$ $|a - b| \geq ||a| - |b||$

$|x - 3| = 5 \implies x - 3 = 5 \text{ or } x - 3 = -5 \implies x = 8 \text{ or } x = -2$

$|x - a| < b \implies a - b < x < a + b \quad (b > 0)$ $|x - a| > b \implies x < a - b \text{ or } x > a + b \quad (b > 0)$

$|x - a| + |y - b| = c$ defines a diamond (rotated square) centred at $(a, b)$.

$|x| + |y| = 4$ defines a diamond with vertices at $(\pm 4, 0)$ and $(0, \pm 4)$.

The domain of a function is the set of all valid inputs; the range is the set of all outputs.

Example: Find the domain and range of $f(x) = \sqrt{4 - x^2}$.

Domain: $4 - x^2 \geq 0 \implies x^2 \leq 4 \implies -2 \leq x \leq 2$.

Range: Since $x^2 \geq 0$, we have $\sqrt{4 - x^2} \leq \sqrt{4} = 2$ and $0 \leq f(x)$.

Therefore, range is $0 \leq f(x) \leq 2$, i.e., $[0, 2]$.

Given $f(x)$ and $g(x)$, the composite function $fg(x) = f(g(x))$.

Example: If $f(x) = 2x + 1$ and $g(x) = x^2$, then:

$fg(x) = f(g(x)) = f(x^2) = 2x^2 + 1$ $gf(x) = g(f(x)) = g(2x + 1) = (2x + 1)^2 = 4x^2 + 4x + 1$

Note that $fg(x) \neq gf(x)$ except when $f$ and $g$ commute.

A function $f$ has an inverse $f^{-1}$ if and only if $f$ is one-to-one (bijective).

$f(f^{-1}(x)) = x \quad \text{and} \quad f^{-1}(f(x)) = x$

Example: If $f(x) = \frac{2x + 3}{x - 1}$, find $f^{-1}(x)$.

Let $y = \frac{2x + 3}{x - 1}$. Solving for $x$:

$y(x - 1) = 2x + 3$ $xy - y = 2x + 3$ $x(y - 2) = y + 3$ $x = \frac{y + 3}{y - 2}$

Therefore, $f^{-1}(x) = \frac{x + 3}{x - 2}$, domain $x \neq 2$.

Writing $f(x) = ax^2 + bx + c$ in vertex form by completing the square:

$f(x) = a\left(x + \frac{b}{2a}\right)^2 + \frac{4ac - b^2}{4a}$

The vertex is at $\left(-\frac{b}{2a}, \frac{4ac - b^2}{4a}\right)$.

$a_n = a_1 + (n-1)d$ $S_n = \frac{n}{2}(a_1 + a_n) = \frac{n}{2}(2a_1 + (n-1)d)$

$a_n = a_1 r^{n-1}$ $S_n = \frac{a_1(1 - r^n)}{1 - r} \quad (r \neq 1)$

For a convergent geometric series (i.e., $|r| < 1$):

$S_\infty = \frac{a_1}{1 - r}$

$\sum_{r=1}^{n} ar^{r-1} = \frac{a(1 - r^n)}{1 - r}$

Example: Evaluate $\sum_{r=1}^{\infty} 3\left(\frac{1}{4}\right)^{r-1}$.

$S_\infty = \frac{3}{1 - \frac{1}{4}} = \frac{3}{\frac{3}{4}} = 4$

For certain sequences, the sum can be found using the method of differences.

Example: Find $\sum_{r=1}^{n} r(r+1)$.

Note that $r(r+1) = r^2 + r$, so:

$\sum_{r=1}^{n} r(r+1) = \sum_{r=1}^{n} r^2 + \sum_{r=1}^{n} r = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$

$= \frac{n(n+1)}{6}(2n + 1 + 3) = \frac{n(n+1)(n+2)}{3}$

Alternatively, using the method of differences:

$\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}$

$\sum_{r=1}^{n} \frac{1}{r(r+1)} = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right) = 1 - \frac{1}{n+1} = \frac{n}{n+1}$

$\lim_{x \to a} f(x) = L$

means that $f(x)$ gets arbitrarily close to $L$ as $x$ gets arbitrarily close to $a$.

$\lim_{x \to a} [f(x) \pm g(x)] = \lim_{x \to a} f(x) \pm \lim_{x \to a} g(x)$ $\lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x)$ $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} \quad \text{(provided } \lim g(x) \neq 0\text{)}$

$\lim_{x \to 0} \frac{\sin x}{x} = 1$ $\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$ $\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e$

$\lim_{x \to \infty} \frac{1}{x^k} = 0 \quad (k > 0)$

For rational functions, divide numerator and denominator by the highest power of $x$:

$\lim_{x \to \infty} \frac{3x^2 + 2x - 1}{5x^2 - x + 3} = \lim_{x \to \infty} \frac{3 + \frac{2}{x} - \frac{1}{x^2}}{5 - \frac{1}{x} + \frac{3}{x^2}} = \frac{3}{5}$

• Power rule: $\frac{d}{dx}(x^n) = nx^{n-1}$
• Chain rule: $\frac{d}{dx}[f(g(x))] = f"(g(x)) \cdot g'(x)$
• Product rule: $\frac{d}{dx}[f(x) \cdot g(x)] = f'(x)g(x) + f(x)g'(x)$
• Quotient rule: $\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$

When $y$ is defined implicitly as a function of $x$, differentiate both sides with respect to $x$ and solve for $\frac{dy}{dx}$.

Example: Find $\frac{dy}{dx}$ given $x^2 + y^2 = 25$.

$2x + 2y\frac{dy}{dx} = 0$ $\frac{dy}{dx} = -\frac{x}{y}$

Example: A spherical balloon is being inflated. Find the rate of change of the radius when the radius is 5 cm, given that air is being pumped in at $100\pi$ cm$^3$/s.

Volume: $V = \frac{4}{3}\pi r^3$.

$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$ $100\pi = 4\pi(25)\frac{dr}{dt}$ $\frac{dr}{dt} = \frac{100\pi}{100\pi} = 1 \text{ cm/s}$

The second derivative $\frac{d^2y}{dx^2}$ represents the rate of change of the gradient. It is used to determine concavity:

• $\frac{d^2y}{dx^2} > 0$: Concave upward (minimum)
• $\frac{d^2y}{dx^2} < 0$: Concave downward (maximum)

To find and classify stationary points:

1. Find $\frac{dy}{dx}$ and set it to zero
2. Solve for $x$ to find the $x$-coordinates of stationary points
3. Find $\frac{d^2y}{dx^2}$ and substitute the $x$-values
4. If $\frac{d^2y}{dx^2} > 0$: minimum; if $< 0$: maximum; if $= 0$: test fails (use first derivative test or higher derivatives)

• Integration by substitution: For integrals of the form $\int f(g(x)) \cdot g'(x) \, dx$, let $u = g(x)$

Example: $\int 2x\sqrt{x^2 + 1} \, dx$. Let $u = x^2 + 1$, $du = 2x \, dx$.

$\int \sqrt{u} \, du = \frac{2}{3}u^{3/2} + C = \frac{2}{3}(x^2 + 1)^{3/2} + C$

• Integration by parts: $\int u \, dv = uv - \int v \, du$

Example: $\int x e^x \, dx$. Let $u = x$, $dv = e^x \, dx$. Then $du = dx$, $v = e^x$.

$\int x e^x \, dx = x e^x - \int e^x \, dx = x e^x - e^x + C = (x - 1)e^x + C$

$\int_a^b f(x) \, dx = \left[F(x)\right]_a^b = F(b) - F(a)$

• Area between a curve and the x-axis: $A = \int_a^b |f(x)| \, dx$
• Area between two curves: $A = \int_a^b |f(x) - g(x)| \, dx$

• Rotation about the x-axis: $V = \pi \int_a^b [f(x)]^2 \, dx$
• Rotation about the y-axis: $V = \pi \int_c^d [f(y)]^2 \, dy$

Example: Find the volume when the region bounded by $y = \sqrt{x}$, $x = 4$, and $y = 0$ is rotated about the x-axis.

$V = \pi \int_0^4 (\sqrt{x})^2 \, dx = \pi \int_0^4 x \, dx = \pi \left[\frac{x^2}{2}\right]_0^4 = 8\pi$

A first-order separable differential equation has the form:

$\frac{dy}{dx} = f(x)g(y)$

Separating variables: $\frac{dy}{g(y)} = f(x) \, dx$.

Integrate both sides and solve for $y$.

Example: Solve $\frac{dy}{dx} = \frac{x}{y}$, given $y = 2$ when $x = 1$.

$y \, dy = x \, dx$ $\int y \, dy = \int x \, dx$ $\frac{y^2}{2} = \frac{x^2}{2} + C$

Using $y = 2$ when $x = 1$: $2 = \frac{1}{2} + C$, so $C = \frac{3}{2}$.

$y^2 = x^2 + 3$

• Forgetting to reverse the inequality sign when multiplying or dividing by a negative number
• Incorrectly finding the domain of composite functions
• Confusing $fg(x)$ notation
• Forgetting to add $+C$ for indefinite integrals
• Not checking whether $\frac{d^2y}{dx^2} = 0$ requires a first derivative test
• Errors in substitution for integration by substitution (forgetting to change the limits for definite integrals or to convert $dx$ to $du$)

flowchart TD
    A[1_M1_Algebra And Calculus] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Mathematical thinking is about abstraction and pattern recognition. Numbers, shapes, and equations are tools for modelling reality - from calculating interest to predicting weather. The beauty of mathematics is that once a pattern is discovered, it can be applied universally. This connects algebra, geometry, and calculus into a coherent framework for understanding quantity, space, and change.

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