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 $h(x) = e^x$. 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: $\displaystyle\lim_{z \to 8} \frac{2z^2 - 17z + 8}{8 - z} = \boxed{-15}$