Calculus

Given: The table of values of $x$ and $y$.

Concept: A relation is a function if and only if every input (value of $x$) has exactly one output (value of $y$).

Working: Looking at the table, the input $x = 1$ appears twice with two different outputs: $y = 0$ and $y = 5$.

Since one input ($x = 1$) maps to two different outputs, this violates the definition of a function.

Conclusion: $y$ is NOT a function of $x$.

Given: The table of values of $x$ and $y$.

Concept: A relation is a function if and only if every input (value of $x$) has exactly one output (value of $y$). Multiple inputs can share the same output — that is perfectly allowed.

Working: Each value of $x$ (namely $-3, -2, -1, 2, 3$) maps to exactly one output $y = 4$. No input is repeated with a different output.

Conclusion: $y$ IS a function of $x$. (It is a constant function $f(x) = 4$.)

Given: $f(x) = x + 1$ and $g(x) = x^2 - 2x + 5$.

Concept: The sum of two functions is defined as $(f + g)(x) = f(x) + g(x)$.

Working:
$$(f + g)(x) = f(x) + g(x)$$
$$= (x + 1) + (x^2 - 2x + 5)$$
$$= x^2 + x - 2x + 1 + 5$$
$$= x^2 - x + 6$$

Answer: $$(f + g)(x) = x^2 - x + 6$$

Graphs: Plot the three functions using GeoGebra or any graphing tool:
- $f(x) = x + 1$ is a straight line with slope 1 and $y$-intercept 1.
- $g(x) = x^2 - 2x + 5$ is an upward-opening parabola.
- $(f+g)(x) = x^2 - x + 6$ is also an upward-opening parabola, shifted compared to $g(x)$.

Instructions: Open GeoGebra Graphing Calculator at https://www.geogebra.org/graphing and type each function in the input bar.

a) $f(x) = x^2$:
- This is an upward-opening parabola with vertex at the origin $(0, 0)$.
- Domain: all real numbers $\mathbb{R}$; Range: $[0, \infty)$.
- The graph is symmetric about the $y$-axis.

b) $f(x) = x^3$:
- This is a cubic curve passing through the origin.
- Domain: $\mathbb{R}$; Range: $\mathbb{R}$.
- The graph is symmetric about the origin (odd function).

c) $f(x) = \dfrac{1}{x}$:
- This is a rectangular hyperbola with two branches in the first and third quadrants.
- Domain: all real numbers except $x = 0$; Range: all real numbers except $0$.
- The $x$-axis and $y$-axis are asymptotes.

*(Graphs to be plotted using GeoGebra as directed.)*

Activity-based question. Open the given GeoGebra applet link in a browser. Interact with the applet to observe:
- The domain is the set of all permissible input values ($x$-values) for the function.
- The range is the set of all output values ($y$-values) produced by the function for inputs in the domain.

Observe how changing the domain affects the range in the applet.

Given: $f(x) = \cos x$ and $f(x) = \tan x$.

For $f(x) = \cos x$:
- The cosine function oscillates between $-1$ and $1$ for all real $x$.
- Domain: $\mathbb{R}$ (all real numbers).
- Range: $[-1,\ 1]$.

For $f(x) = \tan x$:
- The tangent function is undefined at $x = \dfrac{\pi}{2} + n\pi$, $n \in \mathbb{Z}$.
- Between consecutive asymptotes, $\tan x$ takes all real values.
- Domain: $\mathbb{R} \setminus \left\{\dfrac{\pi}{2} + n\pi : n \in \mathbb{Z}\right\}$.
- Range: $\mathbb{R}$ (all real numbers).

*(Graphs to be plotted using a spreadsheet as directed.)*

Given: Three exponential functions $h(x) = e^x$, $g(x) = 10^x$, and $f(x) = 2^x$.

Concept: For exponential functions $a^x$, a larger base $a$ means the function grows faster (steeper graph for x > 0) and falls faster for x < 0.

Comparison of bases: 2 < e \approx 2.718 < 10.

Identification:
- The steepest (fastest growing) graph corresponds to $g(x) = 10^x$ (largest base).
- The least steep graph corresponds to $f(x) = 2^x$ (smallest base).
- The middle graph corresponds to $h(x) = e^x$.

*(The specific colour assignment depends on the GeoGebra plot shown in the figure, which is not visible in the OCR. Students should match the steepness of each curve to the above description to identify the colours.)*

Given: $\displaystyle\lim_{x \to 2} (8 - 3x + 12x^2)$

Concept: For a polynomial function, the limit as $x \to a$ is simply the value of the polynomial at $x = a$ (direct substitution).

Working:
$$\lim_{x \to 2} (8 - 3x + 12x^2) = 8 - 3(2) + 12(2)^2$$
$$= 8 - 6 + 12 \times 4$$
$$= 8 - 6 + 48$$
$$= 50$$

Answer: $\displaystyle\lim_{x \to 2} (8 - 3x + 12x^2) = \boxed{50}$

Given: $\displaystyle\lim_{z \to 8} \frac{2z^2 - 17z + 8}{8 - z}$

Concept: Direct substitution gives $\dfrac{0}{0}$ (indeterminate form), so we factorise the numerator.

Working — Factorise the numerator:
$$2z^2 - 17z + 8$$
We look for two numbers whose product is $2 \times 8 = 16$ and whose sum is $-17$: these are $-16$ and $-1$.
$$2z^2 - 16z - z + 8 = 2z(z - 8) - 1(z - 8) = (2z - 1)(z - 8)$$

Substituting:
$$\lim_{z \to 8} \frac{(2z-1)(z-8)}{8-z} = \lim_{z \to 8} \frac{(2z-1)(z-8)}{-(z-8)}$$

Cancel $(z - 8)$ (valid since $z \neq 8$ in the limit process):
$$= \lim_{z \to 8} \frac{2z - 1}{-1} = \frac{2(8) - 1}{-1} = \frac{15}{-1} = -15$$

Answer: $\displaystyle\lim_{z \to 8} \frac{2z^2 - 17z + 8}{8 - z} = \boxed{-15}$

*(Note: The graph is not visible in the OCR, but based on the solutions provided in the textbook, the answers are as follows.)*

Given: A graph of a function near $x = 3$.

(i) Left-hand limit (as $x \to 3^-$):
$$\lim_{x \to 3^-} f(x) = -1$$
The function approaches $y = -1$ from the left.

(ii) Right-hand limit (as $x \to 3^+$):
$$\lim_{x \to 3^+} f(x) = 2$$
The function approaches $y = 2$ from the right.

(iii) Actual value at $x = 3$:
The actual value of the function at $x = 3$ is $f(3) = 2$.

Conclusion: Since the left-hand limit $(-1) \neq$ right-hand limit $(2)$, the limit of the function at $x = 3$ does not exist. Therefore, the function is not continuous at $x = 3$.

Discussion/Reflection:

(i) $f(x) = c$ (constant function):
For any point $a$, $\lim_{x \to a} c = c = f(a)$. So LHL = RHL = $f(a)$. Hence continuous everywhere.

(ii) $f(x) = x^n$, $n \in \mathbb{N}$:
For any point $a$, $\lim_{x \to a} x^n = a^n = f(a)$ by direct substitution. Hence continuous on $\mathbb{R}$.

(iii) $\sin x$ and $\cos x$:
Both are defined for all real $x$ and their limits equal their values at every point. Hence continuous on $\mathbb{R}$.

(iv) $f(x) = |x|$:
At $x = 0$: LHL $= \lim_{x \to 0^-}(-x) = 0$, RHL $= \lim_{x \to 0^+}(x) = 0$, and $f(0) = 0$. So continuous at $x = 0$ and clearly continuous elsewhere. Hence continuous on $\mathbb{R}$.

(v) Polynomial functions:
Every polynomial $p(x) = a_0x^n + \cdots + a_n$ satisfies $\lim_{x \to a} p(x) = p(a)$ for all $a \in \mathbb{R}$ (by direct substitution). Hence always continuous.

Given: Area of square $= y = x^2$, where $x$ is the side length. At a given moment, $x = a$.

Concept: The instantaneous rate of change of area with respect to side length is the derivative $\dfrac{dy}{dx}$ evaluated at $x = a$.

Finding the derivative using first principle:
$$\frac{dy}{dx} = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h}$$
$$= \lim_{h \to 0} \frac{2xh + h^2}{h} = \lim_{h \to 0} (2x + h) = 2x$$

At $x = a$:
$$\left.\frac{dy}{dx}\right|_{x=a} = 2a$$

Conclusion: The tendency (instantaneous rate of change) of the area at the moment when the side is $a$ is $\mathbf{2a}$ square units per unit length.

Given: Area of a circle $A = \pi r^2$. Find $\dfrac{dA}{dr}$ at $r = 5$ cm.

Concept: Rate of change of area with respect to radius = $\dfrac{dA}{dr}$.

Working:
$$A = \pi r^2$$
$$\frac{dA}{dr} = 2\pi r$$

At $r = 5$ cm:
$$\left.\frac{dA}{dr}\right|_{r=5} = 2\pi (5) = 10\pi \text{ cm}^2/\text{cm}$$

Answer: The rate of change of area with respect to radius when $r = 5$ cm is $\mathbf{10\pi}$ cm²/cm $\approx 31.4$ cm²/cm.

Given:
- Volume of cube: $V = x^3$, where $x$ = edge length.
- $\dfrac{dV}{dt} = 9$ cm³/s.
- Find $\dfrac{dS}{dt}$ when $x = 10$ cm.

Step 1: Relate $V$ and $x$.
$$V = x^3 \implies \frac{dV}{dt} = 3x^2 \frac{dx}{dt}$$
$$9 = 3(10)^2 \frac{dx}{dt} = 300\frac{dx}{dt}$$
$$\frac{dx}{dt} = \frac{9}{300} = \frac{3}{100} \text{ cm/s}$$

Step 2: Surface area of cube.
$$S = 6x^2 \implies \frac{dS}{dt} = 12x \frac{dx}{dt}$$

Step 3: Substitute $x = 10$ and $\dfrac{dx}{dt} = \dfrac{3}{100}$:
$$\frac{dS}{dt} = 12(10) \times \frac{3}{100} = 120 \times \frac{3}{100} = \frac{360}{100} = 3.6 \text{ cm}^2/\text{s}$$

Answer: The surface area is increasing at the rate of $\mathbf{3.6}$ cm²/s.

Given: $f(x) = x^3 + 3x^2 + 1$.

Step 1: Find the derivative.
$$f'(x) = 3x^2 + 6x$$

At $x = -2$:
- Slope: $f'(-2) = 3(4) + 6(-2) = 12 - 12 = 0$
- $y$-coordinate: $f(-2) = (-8) + 3(4) + 1 = -8 + 12 + 1 = 5$
- Equation of tangent: $y = 0(x - (-2)) + 5 \implies y = 5$

At $x = 1$:
- Slope: $f'(1) = 3(1) + 6(1) = 9$
- $y$-coordinate: $f(1) = 1 + 3 + 1 = 5$
- Equation of tangent: $y = 9(x - 1) + 5 = 9x - 9 + 5 \implies y = 9x - 4$

*(Use GeoGebra to visually verify these tangent lines on the graph of $f(x)$.)*

Given: $y = f(x) = x^3 - 2x^2 + x - 3$. Find the tangent at $x = 1$.

Step 1: Find the derivative $f'(x)$.
$$f'(x) = 3x^2 - 4x + 1$$

Step 2: Find the slope at $x = 1$.
$$m = f'(1) = 3(1)^2 - 4(1) + 1 = 3 - 4 + 1 = 0$$

Step 3: Find the $y$-coordinate at $x = 1$.
$$b = f(1) = (1)^3 - 2(1)^2 + (1) - 3 = 1 - 2 + 1 - 3 = -3$$

Step 4: Write the equation of the tangent line.
$$y = m(x - a) + b = 0(x - 1) + (-3)$$
$$\boxed{y = -3}$$

Answer: The equation of the tangent to the curve at $x = 1$ is $y = -3$ (a horizontal line).