Functions -- Diagnostic Tests | DSE

Tests edge cases, boundary conditions, and common misconceptions for functions.

Question:

Let $f(x) = \sqrt{x - 1}$ and $g(x) = \dfrac{1}{x + 2}$.

Find the domain of $f \circ g$.

Solution:

We need the range of $g$ to fall within the domain of $f$.

$\mathrm{dom}(g) = \\{x \in \mathbb{R} : x \neq -2\\}$.

$\mathrm{ran}(g) = \\{y \in \mathbb{R} : y \neq 0\\}$ since $g(x) = \dfrac{1}{x+2}$ can take any non-zero real value.

$\mathrm{dom}(f) = \\{x \in \mathbb{R} : x \geq 1\\}$.

For $f \circ g$ to be defined, we need $g(x) \geq 1$:

$\frac{1}{x+2} \geq 1$

Case 1: $x + 2 > 0$ (i.e. $x > -2$):

$1 \geq x + 2 \implies x \leq -1$

Combined with $x > -2$: $-2 < x \leq -1$.

Case 2: $x + 2 < 0$ (i.e. $x < -2$):

$1 \leq x + 2 \implies x \geq -1$

This contradicts $x < -2$. No solutions in this case.

Therefore $\mathrm{dom}(f \circ g) = (-2,\; -1]$.

Question:

Let $f(x) = \dfrac{2x + 3}{x - 1}$, $x \neq 1$. Which of the following equals $f^{-1}(5)$?

$\text{(A)}\; \frac{1}{f(5)} \qquad \text{(B)}\; f(5) \qquad \text{(C)}\; \text{Neither}$

Find the correct value.

Solution:

$f^{-1}(5)$ is the value of $x$ such that $f(x) = 5$:

$\frac{2x + 3}{x - 1} = 5$

$2x + 3 = 5x - 5$

$3x = 8$

$x = \frac{8}{3}$

Check option (A): $\dfrac{1}{f(5)} = \dfrac{1}{\frac{13}{4}} = \dfrac{4}{13} \neq \dfrac{8}{3}$.

Check option (B): $f(5) = \dfrac{13}{4} \neq \dfrac{8}{3}$.

The answer is (C) Neither. The inverse function evaluated at a point is NOT the reciprocal of the function at that point, nor is it the function itself. $f^{-1}(5) = \dfrac{8}{3}$.

Question:

The graph of $y = f(x)$ passes through the point $(2, 7)$. Which transformation maps this point to $(5, 7)$?

$\text{(A)}\; y = f(x + 3) \qquad \text{(B)}\; y = f(x - 3)$

Solution:

For $y = f(x - 3)$The graph shifts right by 3 units. The point $(2, 7)$ on $y = f(x)$ moves to $(5, 7)$.

For $y = f(x + 3)$The graph shifts left by 3 units. The point $(2, 7)$ moves to $(-1, 7)$.

The answer is (B).

A common mistake is choosing (A) because ”$+3$ looks like moving in the positive direction.” In fact, replacing $x$ with $x - h$ shifts the graph right by $h$Which is the opposite direction to the sign.

Question:

Let $f(x) = x^2 - 4x + 3$, $x \in \mathbb{R}$.

(a) Show that $f$ is not injective. (b) Restrict the domain so that $f^{-1}$ exists and find $f^{-1}(x)$.

Solution:

(a) $f(0) = 3$ and $f(4) = 3$. Since $f(0) = f(4)$ but $0 \neq 4$, $f$ is not injective.

(b) We need to restrict to a domain where $f$ is strictly monotonic.

$f(x) = (x - 2)^2 - 1$ has vertex at $(2, -1)$Opening upward.

Restricting to $\mathrm{dom}(f) = [2,\; \infty)$ makes $f$ strictly increasing.

For $y = (x - 2)^2 - 1$:

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

$x - 2 = \sqrt{y + 1} \quad (\text{since } x \geq 2)$

$x = 2 + \sqrt{y + 1}$

Therefore $f^{-1}(x) = 2 + \sqrt{x + 1}$ with $\mathrm{dom}(f^{-1}) = [-1,\; \infty)$.

Question:

Let $f(x) = 2x - 1$ and $g(x) = x^2 + 4x + 5$. Find the range of $g \circ f$.

Solution:

$g \circ f = g(f(x)) = (2x - 1)^2 + 4(2x - 1) + 5$.

Expanding:

$= 4x^2 - 4x + 1 + 8x - 4 + 5 = 4x^2 + 4x + 2$

Complete the square:

$= 4\left(x^2 + x\right) + 2 = 4\left(x + \tfrac{1}{2}\right)^2 - 1 + 2 = 4\left(x + \tfrac{1}{2}\right)^2 + 1$

Since $4\left(x + \tfrac{1}{2}\right)^2 \geq 0$ for all $x \in \mathbb{R}$:

$\mathrm{ran}(g \circ f) = [1,\; \infty)$

Tests synthesis of functions with other topics.

Question:

Let $f(x) = ax^2 + bx + c$ where $a > 0$. The function $f$ has a minimum value of $-5$ at $x = 3$. Given that $f(1) = 3$Find $a$, $b$ And $c$ And hence find the range of $f^{-1}$.

Solution:

Since the minimum is $-5$ at $x = 3$We can write:

$f(x) = a(x - 3)^2 - 5$

Using $f(1) = 3$:

$a(1 - 3)^2 - 5 = 3$

$4a - 5 = 3 \implies a = 2$

So $f(x) = 2(x - 3)^2 - 5 = 2x^2 - 12x + 13$.

Therefore a = 2$$b = -12$$c = 13.

Since $f$ has minimum value $-5$ and opens upward, $\mathrm{ran}(f) = [-5,\; \infty)$.

Therefore $\mathrm{dom}(f^{-1}) = [-5,\; \infty)$ And so $\mathrm{ran}(f^{-1}) = [3,\; \infty)$ (the restricted domain where $f$ is injective, to the right of the vertex).

Question:

Let f(x) = \log_2(x + 1)$$x > -1. Find $f^{-1}(x)$ and solve $f(x) = f^{-1}(x)$.

Solution:

$y = \log_2(x + 1) \implies 2^y = x + 1 \implies x = 2^y - 1$.

So $f^{-1}(x) = 2^x - 1$ with $\mathrm{dom}(f^{-1}) = \mathbb{R}$.

Solving $f(x) = f^{-1}(x)$:

$\log_2(x + 1) = 2^x - 1$

Let $y = x + 1$ (so $y > 0$):

$\log_2 y = 2^{y-1} - 1$

By inspection: $y = 2$ gives $\log_2 2 = 1$ and $2^{1} - 1 = 1$. Check.

So $x + 1 = 2 \implies x = 1$.

Also $y = 1$ gives $\log_2 1 = 0$ and $2^{0} - 1 = 0$. Check.

So $x + 1 = 1 \implies x = 0$.

The solutions are $x = 0$ and $x = 1$.

Question:

The function f(x) = \dfrac{k}{x}$$x > 0Represents a rectangular hyperbola. The line $y = mx + c$ is tangent to the curve at the point $(2,\; 4)$. Find k$$m And $c$.

Solution:

Since $(2, 4)$ lies on the hyperbola:

$4 = \frac{k}{2} \implies k = 8$

So $f(x) = \dfrac{8}{x}$.

Since $(2, 4)$ lies on the tangent line:

4 = 2m + c \tag{1}

The tangent has the same gradient as the curve at $x = 2$:

$f"(x) = -\frac{8}{x^2} \implies f'(2) = -\frac{8}{4} = -2$

So $m = -2$.

From equation (1): $c = 4 - 2(-2) = 8$.

Therefore k = 8$$m = -2$$c = 8 And the tangent line is $y = -2x + 8$.

Question:

Find the domain and range of $f(x) = \dfrac{2}{x^2 + 1}$.

Solution:

Domain: $x^2 + 1 \neq 0$ for all real $x$ (since $x^2 \geq 0$).

$\mathrm{dom}(f) = \mathbb{R}$

Range: $x^2 + 1 \geq 1$ for all $x$ So $0 < \dfrac{2}{x^2 + 1} \leq 2$.

Maximum value $2$ occurs at $x = 0$. The function approaches $0$ as $|x| \to \infty$.

$\mathrm{ran}(f) = (0,\; 2]$

Question:

Let $f(x) = 2x + 1$ and $g(x) = x^2 - 3$. Find:

(a) $f \circ g(2)$ (b) $g \circ f(2)$ (c) $f \circ f(x)$

Solution:

(a) $g(2) = 4 - 3 = 1$. $f \circ g(2) = f(1) = 3$.

(b) $f(2) = 5$. $g \circ f(2) = g(5) = 25 - 3 = 22$.

Note: $f \circ g(2) \neq g \circ f(2)$Confirming that composition is not commutative.

(c) $f \circ f(x) = f(f(x)) = f(2x + 1) = 2(2x + 1) + 1 = 4x + 3$.

Question:

Let $f(x) = 2x^2 + 4x - 1$ for $x \geq -1$. Find $f^{-1}(x)$.

Solution:

$y = 2x^2 + 4x - 1 = 2(x^2 + 2x) - 1 = 2(x + 1)^2 - 2 - 1 = 2(x + 1)^2 - 3$.

For $x \geq -1$: $y + 3 = 2(x + 1)^2 \geq 0$.

$x + 1 = \sqrt{\frac{y + 3}{2}}$

$x = \sqrt{\frac{y + 3}{2}} - 1$

$f^{-1}(x) = \sqrt{\frac{x + 3}{2}} - 1$

$\mathrm{dom}(f^{-1}) = \mathrm{ran}(f) = [-3,\; \infty)$.

$\mathrm{ran}(f^{-1}) = \mathrm{dom}(f) = [-1,\; \infty)$.

Question:

The graph of $y = f(x)$ passes through $(2, 5)$ and has a minimum at $(3, 1)$. State the corresponding points on the graph of:

(a) $y = f(2x)$ (b) $y = -f(x) + 3$ (c) $y = f(x - 1) + 2$

Solution:

(a) $y = f(2x)$: horizontal compression by factor $\dfrac{1}{2}$.

$(2, 5) \to (1, 5)$ and $(3, 1) \to \left(\dfrac{3}{2},\; 1\right)$.

(b) $y = -f(x) + 3$: reflection in $x$-axis, then shift up 3.

$(2, 5) \to (2, -5) \to (2, -2)$ and $(3, 1) \to (3, -1) \to (3, 2)$.

(c) $y = f(x - 1) + 2$: shift right 1, then up 2.

$(2, 5) \to (3, 7)$ and $(3, 1) \to (4, 3)$.

Question:

Define $f(x) = \begin{cases} x^2 & \text{if } x < 0 \\ 2x + 1 & \text{if } 0 \leq x \leq 3 \\ 7 & \text{if } x > 3 \end{cases}$.

Find f(-2)$$f(0)$$f(3) And $f(5)$.

Solution:

$f(-2) = (-2)^2 = 4$.

$f(0) = 2(0) + 1 = 1$.

$f(3) = 2(3) + 1 = 7$.

$f(5) = 7$.

Question:

Determine whether each function is even, odd, or neither:

(a) $f(x) = x^3 - x$ (b) $g(x) = x^4 + 2x^2$ (c) $h(x) = x^3 + 1$

Solution:

(a) $f(-x) = (-x)^3 - (-x) = -x^3 + x = -(x^3 - x) = -f(x)$. Odd.

(b) $g(-x) = (-x)^4 + 2(-x)^2 = x^4 + 2x^2 = g(x)$. Even.

(c) $h(-x) = (-x)^3 + 1 = -x^3 + 1 \neq h(x)$ and $h(-x) \neq -h(x)$. Neither.

Question:

Let $f(x) = \sqrt{x}$ and $g(x) = x - 4$. Find $f \circ g$ and its domain.

Solution:

$f \circ g(x) = f(g(x)) = f(x - 4) = \sqrt{x - 4}$.

Domain of $f \circ g$: we need $x - 4 \geq 0$I.e. $x \geq 4$.

$\mathrm{dom}(f \circ g) = [4,\; \infty)$

Note: $g(x) = x - 4$ is defined for all $x \in \mathbb{R}$ But the range of $g$ must fall within the domain of $f$ (which is $[0, \infty)$), so $g(x) \geq 0 \implies x \geq 4$.

Question:

Determine whether $f(x) = \dfrac{2x + 3}{x - 1}$ is injective.

Solution:

Suppose $f(a) = f(b)$:

$\frac{2a + 3}{a - 1} = \frac{2b + 3}{b - 1}$

$(2a + 3)(b - 1) = (2b + 3)(a - 1)$

$2ab - 2a + 3b - 3 = 2ab - 2b + 3a - 3$

$-2a + 3b = -2b + 3a$

$5b = 5a$

$a = b$

Since $f(a) = f(b) \implies a = b$The function is injective.

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

A machine with input and output: A function is like a vending machine — put in a specific item (input), get out a specific snack (output). Composition is chaining machines: the output of one becomes the input of the next. The inverse is the undo button.

Why it matters: Functions model everything from bacterial growth to financial returns. Understanding domain, range, and composition lets you predict system behavior and reverse-engineer processes.

The key insight: The inverse function “undoes” the original — f⁻¹(f(x)) = x — but only works when the function is one-to-one.

1. Confusing $f^{-1}$ with the reciprocal $\dfrac{1}{f}$. The notation $f^{-1}$ denotes the inverse function, NOT the reciprocal. $f^{-1}(x)$ is the value of $y$ such that $f(y) = x$Which is completely different from $\dfrac{1}{f(x)}$.

2. Incorrect domain of composite functions. The domain of $f \circ g$ is NOT $\mathrm{dom}(g)$. It is the set of all $x$ in $\mathrm{dom}(g)$ such that $g(x) \in \mathrm{dom}(f)$. You must check both conditions.

3. Wrong direction for horizontal shifts. $f(x - h)$ shifts the graph RIGHT by $h$ units, not left. The transformation is counterintuitive: replacing $x$ with $x - h$ moves the graph in the positive $x$-direction.

4. Forgetting to restrict the domain when finding the inverse of a non-injective function. If $f$ is not one-to-one on its entire domain, you must restrict the domain before finding the inverse. Always state the restricted domain explicitly.

5. Assuming $f \circ g = g \circ f$. Function composition is generally NOT commutative. Always compute $f \circ g$ and $g \circ f$ separately unless you have proven they are equal.

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

(a) Find $f^{-1}(x)$. (3 marks) (b) Find the domain and range of $f^{-1}$. (2 marks) (c) Solve $f(x) = x$. (3 marks)

Solution:

(a) $y = \dfrac{3x - 1}{x + 2}$.

$y(x + 2) = 3x - 1 \implies yx + 2y = 3x - 1 \implies yx - 3x = -1 - 2y$.

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

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

$f^{-1}(x) = \frac{2x + 1}{3 - x}$

(b) $\mathrm{dom}(f^{-1}) = \mathrm{ran}(f)$. Since $f(x) = \dfrac{3x - 1}{x + 2} = 3 - \dfrac{7}{x + 2}$ and $\dfrac{7}{x+2}$ takes all non-zero real values, $\mathrm{ran}(f) = \mathbb{R} \setminus \{3\}$.

$\mathrm{dom}(f^{-1}) = \{x \in \mathbb{R} : x \neq 3\}$.

$\mathrm{ran}(f^{-1}) = \mathrm{dom}(f) = \{x \in \mathbb{R} : x \neq -2\}$.

(c) $\dfrac{3x - 1}{x + 2} = x \implies 3x - 1 = x^2 + 2x \implies x^2 - x + 1 = 0$.

$\Delta = 1 - 4 = -3 < 0$. No real solutions.

Let $f(x) = x^2 - 6x + 5$ and $g(x) = 2x - 3$.

(a) Express $f(x)$ in the form $(x - a)^2 + b$. (2 marks) (b) Find the range of $f$. (1 mark) (c) Find $f \circ g(x)$ and $g \circ f(x)$. (4 marks) (d) Solve $f \circ g(x) = 0$. (2 marks)

Solution:

(a) $f(x) = (x - 3)^2 - 9 + 5 = (x - 3)^2 - 4$.

(b) Since $(x - 3)^2 \geq 0$: $\mathrm{ran}(f) = [-4,\; \infty)$.

(c) $f \circ g(x) = f(2x - 3) = (2x - 3 - 3)^2 - 4 = (2x - 6)^2 - 4 = 4x^2 - 24x + 36 - 4 = 4x^2 - 24x + 32$.

$g \circ f(x) = g(x^2 - 6x + 5) = 2(x^2 - 6x + 5) - 3 = 2x^2 - 12x + 7$.

(d) $4x^2 - 24x + 32 = 0 \implies x^2 - 6x + 8 = 0 \implies (x-2)(x-4) = 0 \implies x = 2$ or $x = 4$.

The function $f$ is defined by $f(x) = \dfrac{1}{x^2 - 4}$.

(a) Find the domain of $f$. (1 mark) (b) Find the range of $f$. (3 marks) (c) Solve $f(x) = \dfrac{1}{5}$. (2 marks)

Solution:

(a) $x^2 - 4 \neq 0 \implies x \neq \pm 2$.

$\mathrm{dom}(f) = \{x \in \mathbb{R} : x \neq -2 \text{ and } x \neq 2\}$

(b) Let $y = \dfrac{1}{x^2 - 4}$. Then $x^2 - 4 = \dfrac{1}{y}$ So $x^2 = 4 + \dfrac{1}{y} = \dfrac{4y + 1}{y}$.

For $x^2 \geq 0$: $\dfrac{4y + 1}{y} \geq 0$.

Critical values: $y = 0$ (asymptote) and $y = -\dfrac{1}{4}$.

$\dfrac{4y + 1}{y} \geq 0 \implies y < -\dfrac{1}{4}$ or $y > 0$.

$\mathrm{ran}(f) = \left(-\infty,\; -\dfrac{1}{4}\right) \cup (0,\; \infty)$

(c) $\dfrac{1}{x^2 - 4} = \dfrac{1}{5} \implies x^2 - 4 = 5 \implies x^2 = 9 \implies x = \pm 3$.

Let $f(x) = 2^x$ and $g(x) = \log_2 x$.

(a) Find $f \circ g(x)$ and simplify. (2 marks) (b) Find $g \circ f(x)$ and simplify. (2 marks) (c) Explain the relationship between $f$ and $g$. (1 mark)

Solution:

(a) $f \circ g(x) = f(\log_2 x) = 2^{\log_2 x} = x$For $x > 0$.

(b) $g \circ f(x) = g(2^x) = \log_2(2^x) = x$For all $x \in \mathbb{R}$.

(c) $f$ and $g$ are inverse functions of each other. $f \circ g = \mathrm{id}$ on $(0, \infty)$ and $g \circ f = \mathrm{id}$ on $\mathbb{R}$.

The graph of $y = f(x)$ is shown. It passes through (-2, 0)$$(0, 4)$$(2, 0) And has a maximum at $(0, 4)$.

(a) Sketch the graph of $y = f(x + 1)$. (2 marks) (b) Sketch the graph of $y = f(-x)$. (2 marks) (c) The graph of $y = f(x)$ is transformed to the graph of $y = -f(x) + 2$. Describe this transformation in words. (2 marks)

Solution:

(a) $y = f(x + 1)$ shifts the graph left by 1 unit. New key points: (-3, 0)$$(-1, 4)$$(1, 0). Maximum at $(-1, 4)$.

(b) $y = f(-x)$ reflects the graph in the $y$-axis. New key points: (2, 0)$$(0, 4)$$(-2, 0). Maximum at $(0, 4)$.

(c) $y = -f(x) + 2$: reflect in the $x$-axis (all $y$-values change sign), then translate up by 2 units. New maximum at $(0, -4 + 2) = (0, -2)$. New $x$-intercepts at $(-2, 2)$ and $(2, 2)$ (which are not on the $x$-axis anymore).

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