Adjust the parameters in the graph above to explore the relationships between variables.

A function maps each element of a set (the domain) to exactly one element of another set (the Codomain). If $f(x) = 3x + 2$Then $f$ takes an input $x$ and returns $3x + 2$. The notation $f \colon A \to B$ means $f$ maps from set $A$ to set $B$.

Domain and Range:

The domain of $f$ is the set of all valid inputs. The range is the set of all outputs.

The domain of $f(x) = \dfrac{1}{x - 3}$ is $x \in \mathbb{R}$, $x \neq 3$.

The range of $f(x) = x^2$ is $f(x) \geq 0$ (assuming the domain is all real numbers).

Composite Functions:

If $f(x) = 2x + 1$ and $g(x) = x^2$Then:

$f(g(x)) = f(x^2) = 2x^2 + 1$

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

Note that $f(g(x)) \neq g(f(x))$. Composition is not commutative.

Domain of a composite function: $\operatorname{dom}(f \circ g)$ is the set of all $x$ in $\operatorname{dom}(g)$ such that $g(x) \in \operatorname{dom}(f)$.

Inverse Functions:

The inverse function $f^{-1}$ reverses the effect of $f$. To find $f^{-1}$:

1. Write $y = f(x)$
2. Rearrange to make $x$ the subject
3. Replace $x$ with $f^{-1}(y)$

Example: Find $f^{-1}$ where $f(x) = \dfrac{2x + 3}{x - 1}$.

Let $y = \dfrac{2x + 3}{x - 1}$.

$y(x - 1) = 2x + 3$

$yx - y = 2x + 3$

$yx - 2x = y + 3$

$x(y - 2) = y + 3$

$x = \frac{y + 3}{y - 2}$

Therefore $f^{-1}(x) = \dfrac{x + 3}{x - 2}$.

Verification: $f(f^{-1}(x)) = \dfrac{2 \cdot \frac{x+3}{x-2} + 3}{\frac{x+3}{x-2} - 1} = \dfrac{\frac{2x+6+3x-6}{x-2}}{\frac{x+3-x+2}{x-2}} = \dfrac{5x}{5} = x$. Confirmed.

A quadratic function has the general form $f(x) = ax^2 + bx + c$ where $a \neq 0$.

Completed Square Form:

$f(x) = a(x - p)^2 + q$

Where the vertex is at $(p, q)$. The axis of symmetry is the vertical line $x = p$.

Derivation. Starting from $ax^2 + bx + c$:

$$Ax^2 + bx + c = a\!\left(x^2 + \frac{b}{a}x\right) + c = a\!\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c$$

So $p = -\frac{b}{2a}$ and $q = c - \frac{b^2}{4a}$.

Example: Express $f(x) = 2x^2 - 12x + 7$ in completed square form.

$f(x) = 2(x^2 - 6x) + 7$

$f(x) = 2\left((x - 3)^2 - 9\right) + 7$

$f(x) = 2(x - 3)^2 - 18 + 7$

$f(x) = 2(x - 3)^2 - 11$

The vertex is at $(3, -11)$ and since $a = 2 > 0$The parabola opens upward with a minimum value of $-11$.

The Discriminant:

For $ax^2 + bx + c = 0$The discriminant is $\Delta = b^2 - 4ac$.

Condition	Number of Roots
$\Delta > 0$	Two distinct real roots
$\Delta = 0$	One repeated real root
$\Delta < 0$	No real roots

Proof. The quadratic formula gives $x = \frac{-b \pm \sqrt{\Delta}}{2a}$. If $\Delta > 0$The Square root is real and positive, yielding two distinct values. If $\Delta = 0$Both values are $-\frac{b}{2a}$. If $\Delta < 0$The square root is not real.

Example: Determine the nature of the roots of $3x^2 - 5x + 2 = 0$.

$\Delta = (-5)^2 - 4(3)(2) = 25 - 24 = 1$

Since $\Delta > 0$There are two distinct real roots.

Example: Find the value of $k$ for which $x^2 + kx + 9 = 0$ has equal roots.

$\Delta = k^2 - 36 = 0 \implies k = \pm 6$

Laws of Logarithms:

$\log_a(xy) = \log_a x + \log_a y$

$\log_a\left(\frac{x}{y}\right) = \log_a x - \log_a y$

$\log_a(x^n) = n \log_a x$

$\log_a a = 1, \quad \log_a 1 = 0$

Change of Base:

$$\log_a b = \frac{\log_c b}{\log_c a}$$

Proof of change of base. Let $y = \log_a b$So $a^y = b$. Taking $\log_c$ of both sides: $y \log_c a = \log_c b$Hence $y = \frac{\log_c b}{\log_c a}$.

Example: Solve $3^{2x - 1} = 7$.

$2x - 1 = \log_3 7 = \frac{\ln 7}{\ln 3}$

$2x = 1 + \frac{\ln 7}{\ln 3}$

$$X = \frac{1}{2}\left(1 + \frac{\ln 7}{\ln 3}\right) \approx 1.389$$

Example: Solve $\log_2(x + 3) + \log_2(x - 1) = 4$.

$\log_2((x + 3)(x - 1)) = 4$

$(x + 3)(x - 1) = 16$

$x^2 + 2x - 3 = 16$

$x^2 + 2x - 19 = 0$

$x = \frac{-2 \pm \sqrt{4 + 76}}{2} = \frac{-2 \pm \sqrt{80}}{2} = -1 \pm 2\sqrt{5}$

Since $x + 3 > 0$ and $x - 1 > 0$We need $x > 1$. So $x = -1 + 2\sqrt{5} \approx 3.472$.

Exponential Growth and Decay:

$N(t) = N_0 e^{kt}$

Where $N_0$ is the initial quantity and $k$ is the growth ($k > 0$) or decay ($k < 0$) constant.

Half-life: For decay with half-life $t_{1/2}$:

$$T_{1/2} = \frac{\ln 2}{|k|}$$

Proof of the half-life formula. Set $N(t_{1/2}) = \frac{N_0}{2}$:

$$\frac{N_0}{2} = N_0 e^{kt_{1/2}} \implies \frac{1}{2} = e^{kt_{1/2}} \implies \ln\!\left(\frac{1}{2}\right) = kt_{1/2}$$ $$-\ln 2 = kt_{1/2} \implies t_{1/2} = \frac{-\ln 2}{k} = \frac{\ln 2}{|k|}$$

$\blacksquare$

Example: A radioactive substance has a half-life of 8 days. If a sample initially contains 200 G, how much remains after 25 days?

$$K = \frac{\ln 2}{8} \approx 0.0866$$ N(25) = 200e^{-0.0866 \times 25} = 200e^{-2.165} \approx 200 \times 0.1147 \approx 22.9 \mathrm{ g

A polynomial of degree $n$ has the form:

$$P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$$

Factor Theorem:

$(x - a)$ is a factor of $p(x)$ if and only if $p(a) = 0$.

Remainder Theorem:

When $p(x)$ is divided by $(x - a)$The remainder is $p(a)$.

Proof. By polynomial long division, $p(x) = (x-a)q(x) + r$ for some quotient $q(x)$ and constant Remainder $r$. Setting $x = a$: $p(a) = 0 + r$So $r = p(a)$.

Example: Factorise $p(x) = x^3 - 3x^2 - 4x + 12$.

Try $p(1) = 1 - 3 - 4 + 12 = 6 \neq 0$.

Try $p(2) = 8 - 12 - 8 + 12 = 0$. So $(x - 2)$ is a factor.

$$P(x) = (x - 2)(x^2 - x - 6) = (x - 2)(x - 3)(x + 2)$$

Example: Find the remainder when $p(x) = 2x^3 + x^2 - 5x + 3$ is divided by $(x + 1)$.

By the Remainder Theorem, the remainder is $p(-1)$:

$$P(-1) = 2(-1)^3 + (-1)^2 - 5(-1) + 3 = -2 + 1 + 5 + 3 = 7$$

Example: Given $p(x) = x^3 + 2x^2 - 5x - 6$Show that $(x + 1)$ is a factor and hence fully Factorise $p(x)$.

$p(-1) = -1 + 2 + 5 - 6 = 0$. So $(x + 1)$ is a factor.

By polynomial long division:

$$P(x) = (x + 1)(x^2 + x - 6) = (x + 1)(x + 3)(x - 2)$$

Linear-Quadratic Systems:

Substitute the linear equation into the quadratic. The resulting equation is always a quadratic, Giving at most two solutions.

Example: Solve the system $y = x^2 - 4x + 3$ and $2x + y = 6$.

Substitute: $2x + x^2 - 4x + 3 = 6$

$x^2 - 2x - 3 = 0$

$(x - 3)(x + 1) = 0$

x = 3 \mathrm{ or x = -1

When $x = 3$: $y = 9 - 12 + 3 = 0$. When $x = -1$: $y = 1 + 4 + 3 = 8$.

Solutions: $(3, 0)$ and $(-1, 8)$.

Given $y = f(x)$:

Transformation	Effect
$y = f(x) + c$	Vertical translation by $c$ units up
$y = f(x - c)$	Horizontal translation by $c$ units right
$y = af(x)$	Vertical stretch by factor $a$
$y = f(bx)$	Horizontal stretch by factor $\frac{1}{b}$
$y = -f(x)$	Reflection in the $x$-axis
$y = f(-x)$	Reflection in the $y$-axis

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The derivative of $f(x)$ measures the instantaneous rate of change of $f$ with respect to $x$. Geometrically, it gives the gradient of the tangent to the curve $y = f(x)$.

$$F'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$

Standard Derivatives:

$f(x)$	$f'(x)$
$x^n$	$nx^{n-1}$
$e^{kx}$	$ke^{kx}$
$\ln x$	$\dfrac{1}{x}$
$\sin x$	$\cos x$
$\cos x$	$-\sin x$

Differentiation Rules:

Sum/Difference: $(f \pm g)' = f' \pm g'$

Product Rule: $(fg)' = f'g + fg'$

Quotient Rule: $\left(\dfrac{f}{g}\right)' = \dfrac{f'g - fg'}{g^2}$

Chain Rule: If $y = f(g(x))$Then $\dfrac{dy}{dx} = f'(g(x)) \cdot g'(x)$

Proof of the product rule. Define $u(x) = f(x)g(x)$. Then:

$$U'(x) = \lim_{h \to 0} \frac{f(x+h)g(x+h) - f(x)g(x)}{h}$$

Add and subtract $f(x)g(x+h)$ in the numerator:

$$= \lim_{h \to 0} \frac{f(x+h)g(x+h) - f(x)g(x+h) + f(x)g(x+h) - f(x)g(x)}{h}$$ $$= \lim_{h \to 0} \left[g(x+h) \cdot \frac{f(x+h) - f(x)}{h} + f(x) \cdot \frac{g(x+h) - g(x)}{h}\right]$$ $$= g(x)f'(x) + f(x)g'(x)$$

Example: Differentiate $f(x) = (3x^2 + 1)\sin(2x)$ using the product rule.

Let $u = 3x^2 + 1$, $v = \sin(2x)$.

$u' = 6x$, $v' = 2\cos(2x)$.

$$F'(x) = 6x \sin(2x) + (3x^2 + 1) \cdot 2\cos(2x)$$ $$F'(x) = 6x \sin(2x) + 2(3x^2 + 1)\cos(2x)$$

Example: Differentiate $f(x) = \dfrac{e^{2x}}{x + 1}$.

Let $u = e^{2x}$, $v = x + 1$.

$u' = 2e^{2x}$, $v' = 1$.

$$F'(x) = \frac{2e^{2x}(x + 1) - e^{2x}}{(x + 1)^2} = \frac{e^{2x}(2x + 2 - 1)}{(x + 1)^2} = \frac{e^{2x}(2x + 1)}{(x + 1)^2}$$

Example: Differentiate $f(x) = \dfrac{x^2 + 1}{e^{3x}}$.

Let $u = x^2 + 1$, $v = e^{3x}$.

$u' = 2x$, $v' = 3e^{3x}$.

$$F'(x) = \frac{2x \cdot e^{3x} - (x^2 + 1) \cdot 3e^{3x}}{e^{6x}} = \frac{e^{3x}(2x - 3x^2 - 3)}{e^{6x}} = \frac{2x - 3x^2 - 3}{e^{3x}}$$

Setting $f'(x) = 0$: $2x - 3x^2 - 3 = 0$I.e., $3x^2 - 2x + 3 = 0$. Since $\Delta = 4 - 36 = -32 \lt 0$There are no real stationary points.

Stationary Points:

At a stationary point, $f'(x) = 0$.

• $f'(x)$ changes from positive to negative: local maximum
• $f'(x)$ changes from negative to positive: local minimum
• $f'(x)$ does not change sign: point of inflection

Second Derivative Test:

If $f'(a) = 0$:

• $f''(a) > 0$: local minimum at $x = a$
• $f''(a) < 0$: local maximum at $x = a$
• $f''(a) = 0$: test is inconclusive — use the first derivative test instead

Example: Find the stationary points of $f(x) = x^3 - 6x^2 + 9x + 1$ and determine their nature.

$$F'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x - 1)(x - 3)$$

Stationary points at $x = 1$ and $x = 3$.

$$F''(x) = 6x - 12$$

At $x = 1$: $f''(1) = -6 < 0$So local maximum. $f(1) = 1 - 6 + 9 + 1 = 5$.

At $x = 3$: $f''(3) = 6 > 0$So local minimum. $f(3) = 27 - 54 + 27 + 1 = 1$.

Local maximum at $(1, 5)$Local minimum at $(3, 1)$.

The tangent to $y = f(x)$ at $(a, f(a))$ has equation:

$$Y - f(a) = f'(a)(x - a)$$

The normal is perpendicular to the tangent and has gradient $-\dfrac{1}{f'(a)}$ (provided $f'(a) \ne 0$).

Example: Find the equation of the tangent to $y = x^3 - 3x + 1$ at $x = 2$.

$f(2) = 8 - 6 + 1 = 3$.

$f'(x) = 3x^2 - 3$So $f'(2) = 12 - 3 = 9$.

$$Y - 3 = 9(x - 2) \implies y = 9x - 15$$

Optimisation problems require you to express the quantity to be optimised as a function of a single Variable, then find the stationary points.

Example: A rectangular box with a square base has a volume of 128 \mathrm{ cm^3. The material For the base costs 5 pence per \mathrm{cm^2 and the material for the sides costs 3 pence per \mathrm{cm^2. Find the dimensions that minimise the cost.

Let the base have side length $x$ cm and height $h$ cm.

Volume: $x^2 h = 128$So $h = \dfrac{128}{x^2}$.

Cost: $C = 5x^2 + 4 \times 3xh = 5x^2 + 12x \cdot \dfrac{128}{x^2} = 5x^2 + \dfrac{1536}{x}$.

$$\frac{dC}{dx} = 10x - \frac{1536}{x^2}$$

Setting $\dfrac{dC}{dx} = 0$:

$$10x = \frac{1536}{x^2}$$ $$10x^3 = 1536$$ $$X^3 = 153.6$$ X = \sqrt[3]{153.6} \approx 5.35 \mathrm{ cm H = \frac{128}{5.35^2} \approx 4.47 \mathrm{ cm

Verification: $\frac{d^2C}{dx^2} = 10 + \frac{3072}{x^3} > 0$ for all $x > 0$Confirming a Minimum.

Example: A closed cylindrical can must hold 500 \mathrm{ cm^3 of liquid. Find the dimensions That minimise the surface area.

Let the radius be $r$ and the height be $h$.

Volume: $\pi r^2 h = 500$So $h = \dfrac{500}{\pi r^2}$.

Surface area: $S = 2\pi r^2 + 2\pi rh = 2\pi r^2 + \frac{1000}{r}$.

$$\frac{dS}{dr} = 4\pi r - \frac{1000}{r^2} = 0$$ 4\pi r^3 = 1000 \implies r^3 = \frac{250}{\pi} \implies r = \sqrt[3]{\frac{250}{\pi}} \approx 4.30 \mathrm{ cm H = \frac{500}{\pi \times 4.30^2} \approx 8.60 \mathrm{ cm

Note that $h = 2r$So the optimal can has height equal to diameter. $\frac{d^2S}{dr^2} = 4\pi + \frac{2000}{r^3} > 0$Confirming a minimum.

Integration is the reverse of differentiation.

$$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)$$ $$\int e^{kx} \, dx = \frac{e^{kx}}{k} + C$$

Definite Integration:

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

The Fundamental Theorem of Calculus. If $F'(x) = f(x)$ on $[a, b]$Then $\int_a^b f(x)\,dx = F(b) - F(a)$. This theorem connects the two branches of calculus: Differentiation and integration are inverse operations.

Area Under a Curve:

The area between $y = f(x)$The $x$-axis, $x = a$And $x = b$ is:

$$A = \int_a^b |f(x)| \, dx$$

:::caution If the curve crosses the $x$-axis between $a$ and $b$You must split the integral at Each crossing point and take the absolute value. The integral itself gives the signed area, which Can cancel out. :::

Example: Find the area enclosed by $y = x^2 - 4x + 3$ and the $x$-axis.

Find where the curve crosses the $x$-axis: $x^2 - 4x + 3 = (x-1)(x-3) = 0$So $x = 1$ and $x = 3$.

Since the parabola opens upward with roots at 1 and 3, the curve is below the $x$-axis between these Points.

$$A = \int_1^3 |x^2 - 4x + 3| \, dx = \int_1^3 (4x - x^2 - 3) \, dx$$ $$= \left[2x^2 - \frac{x^3}{3} - 3x\right]_1^3$$ $$= \left(18 - 9 - 9\right) - \left(2 - \frac{1}{3} - 3\right)$$ $$= 0 - \left(-\frac{4}{3}\right) = \frac{4}{3}$$

If $f(x) \ge g(x)$ on $[a, b]$The area between the curves is:

$$A = \int_a^b [f(x) - g(x)] \, dx$$

Example: Find the area between $y = x^2$ and $y = 2x$.

Find intersections: $x^2 = 2x$So $x^2 - 2x = 0$Giving $x = 0$ and $x = 2$.

Between $x = 0$ and $x = 2$, $2x > x^2$.

$$A = \int_0^2 (2x - x^2) \, dx = \left[x^2 - \frac{x^3}{3}\right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}$$

Example: Find the area enclosed by the curves $y = x^3$ and $y = x$.

Find intersections: $x^3 = x$So $x^3 - x = x(x-1)(x+1) = 0$Giving $x = -1, 0, 1$.

Between $x = -1$ and $x = 0$: $x \ge x^3$ (since $x \in [-1, 0]$ means $x \ge x^3$). Between $x = 0$ And $x = 1$: $x^3 \ge x$.

By symmetry of the two regions:

$$A = 2\int_0^1 (x^3 - x)\, dx = 2\left[\frac{x^4}{4} - \frac{x^2}{2}\right]_0^1 = 2\!\left(\frac{1}{4} - \frac{1}{2}\right) = 2\!\left(-\frac{1}{4}\right) = \frac{1}{2}$$\int f(g(x))g'(x) \, dx = \int f(u) \, du \quad \mathrm{where u = g(x)

Example: Evaluate $\int 2x\sqrt{x^2 + 1} \, dx$.

Let $u = x^2 + 1$So $du = 2x\,dx$.

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

Example: Evaluate $\int_0^2 \frac{x}{x^2 + 1}\,dx$.

Let u = x^2 + 1$$du = 2x\,dx. When x = 0$$u = 1. When x = 2$$u = 5.

$$\int_0^2 \frac{x}{x^2 + 1}\,dx = \frac{1}{2}\int_1^5 \frac{1}{u}\,du = \frac{1}{2}\left[\ln u\right]_1^5 = \frac{1}{2}\ln 5$$$$\int u\, dv = uv - \int v\, du$$

Use LIATE (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential) to choose $u$.

Example: Evaluate $\int x e^x \, dx$.

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

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

A first-order differential equation relates a function to its first derivative.

Separable equations. If $\dfrac{dy}{dx} = f(x)g(y)$Separate the variables:

$$\frac{1}{g(y)}\,dy = f(x)\,dx$$

Integrate both sides.

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

Separate: $y\,dy = x\,dx$.

Integrate: $\dfrac{y^2}{2} = \dfrac{x^2}{2} + C$.

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

$y^2 = x^2 + 3$Giving $y = \sqrt{x^2 + 3}$ (taking the positive root since $y = 2 > 0$).

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See the examples integrated throughout the sections above.

1. Forgetting the chain rule: When differentiating $\sin(3x)$The answer is $3\cos(3x)$Not $\cos(3x)$. Every composite function requires the chain rule.

2. Missing the $+C$: Always include the constant of integration for indefinite integrals. Omitting it is equivalent to asserting that you know the particular solution without initial conditions.

3. Sign errors in the product rule: It is $f'g + fg'$Not $f'g - fg'$. The minus sign belongs in the quotient rule.

4. Incorrect domain for logarithms: $\ln(x)$ is only defined for $x > 0$. Always check your solutions satisfy the domain restrictions. When solving $\log_2(x+3)$You need $x > -3$.

5. Confusing $f^{-1}(x)$ with $\dfrac{1}{f(x)}$: The notation $f^{-1}$ denotes the inverse function, not the reciprocal. This is a notation collision that causes persistent confusion.

6. Forgetting absolute values in integrals of $1/x$: $\int \frac{dx}{x} = \ln|x| + C$Not $\ln x + C$.

7. Not checking that optimisation solutions are minima: Always verify with the second derivative test or a sign chart.

8. Forgetting to change the limits when using substitution for definite integrals.

9. Sign error in the quotient rule: It is $\dfrac{f'g - fg'}{g^2}$Not $\dfrac{fg' - f'g}{g^2}$. “Low d-High minus High d-Low.”

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1. Given $f(x) = 2x^2 - 8x + 5$Express $f(x)$ in completed square form and state the coordinates of the vertex.

2. Solve $4^{x+1} = 3^{2x-1}$Giving your answer in terms of natural logarithms.

3. Given $p(x) = x^3 + 2x^2 - 5x - 6$Show that $(x + 1)$ is a factor and hence fully factorise $p(x)$.

4. Differentiate $f(x) = \dfrac{x^2 + 1}{e^{3x}}$ and find the coordinates of any stationary points.

5. Find the area enclosed by the curves $y = x^3$ and $y = x$.

6. A closed cylindrical can must hold 500 \mathrm{ cm^3 of liquid. Find the dimensions that minimise the surface area.

7. Solve $\log_3(x) + \log_3(x - 2) = 1$.

8. Find the equation of the tangent to $y = x^3 - 3x + 1$ at the point where $x = 2$.

9. Given $f(x) = \frac{1}{3}x^3 - x^2 - 3x + 2$Find the intervals on which $f$ is increasing and decreasing.

10. Evaluate $\int_0^2 \frac{x}{x^2 + 1}\,dx$ using substitution.

11. The curve $y = x^2 + px + q$ passes through $(1, 5)$ and has a turning point at $x = 2$. Find $p$ and $q$.

12. Find the area between the curves $y = x^2 + 1$ and $y = 3x + 1$.

13. Solve the differential equation $\dfrac{dy}{dx} = 2xy$ given $y = 5$ when $x = 0$.

14. Find the equation of the normal to $y = x^2 - 4x + 3$ at the point where $x = 4$.

15. A population of bacteria grows according to $P(t) = P_0 e^{0.02t}$. If the initial population is 1000, how long does it take for the population to reach 5000?

16. Find the range of values of $k$ for which $x^2 + 2kx + k^2 + 3 = 0$ has no real roots.

This topic covers the mathematical techniques and concepts related to algebra and calculus, including key theorems, methods, and problem-solving approaches.

Key concepts include:

• quadratic equations and the discriminant
• simultaneous equations
• polynomial division and the factor theorem
• partial fractions
• binomial expansion

Regular practice with a variety of question types is essential to build fluency and confidence in applying these mathematical techniques.