Questions on Functions with Solutions

This page presents a set of carefully selected questions on functions, each followed by a detailed solution. The questions cover key concepts such as the definition of a function, domain and range, evaluation, composition, and graph transformations.


Question 1

Is the graph shown below the graph of a function?

Graph used to test whether it represents a function

Solution

By the vertical line test, a vertical line at \(x = 0\) intersects the graph at two points. Therefore, the graph does not represent a function.


Question 2

Does the equation \[ y^2 + x = 1 \] represent a function \(y\) in terms of \(x\)?

Solution

Solving for \(y\), \[ y^2 = 1 - x \] \[ y = \pm \sqrt{1 - x} \] For a given value of \(x\), there are two possible values of \(y\). Hence, the relation does not define a function.


Question 3

The function \(f\) is defined by \[ f(x) = -2x^2 + 6x - 3 \] Find \(f(-2)\).

Solution

\[ f(-2) = -2(-2)^2 + 6(-2) - 3 = -23 \]


Question 4

The function \(h\) is defined by \[ h(x) = 3x^2 - 7x - 5 \] Find \(h(x - 2)\).

Solution

Substitute \(x - 2\) for \(x\): \[ h(x - 2) = 3(x - 2)^2 - 7(x - 2) - 5 \] \[ = 3(x^2 - 4x + 4) - 7x + 14 - 5 \] \[ = 3x^2 - 19x + 7 \]


Question 5

The functions \(f\) and \(g\) are defined by \[ f(x) = -7x - 5, \qquad g(x) = 10x - 12 \] Find \((f + g)(x)\).

Solution

\[ (f + g)(x) = (-7x - 5) + (10x - 12) = 3x - 17 \]


Question 6

\[ f(x) = \frac{1}{x} + 3x, \qquad g(x) = -\frac{1}{x} + 6x - 4 \] Find \((f + g)(x)\) and its domain.

Solution

\[ (f + g)(x) = 9x - 4 \] The domain excludes \(x = 0\): \[ (-\infty, 0) \cup (0, \infty) \]


Question 7

\[ f(x) = x^2 - 2x + 1, \qquad g(x) = (x - 1)(x + 3) \] Find \((f / g)(x)\) and its domain.

Solution

\[ \frac{f(x)}{g(x)} = \frac{(x - 1)^2}{(x - 1)(x + 3)} = \frac{x - 1}{x + 3} \] Restrictions: \(x \neq -3, 1\) \[ (-\infty,-3)\cup(-3,1)\cup(1,\infty) \]


Question 8

Find the domain of \[ h(x) = \sqrt{x - 2} \]

Solution

\[ x - 2 \ge 0 \Rightarrow x \ge 2 \] Domain: \[ [2, \infty) \]


Question 9

\[ g(x) = \sqrt{-x^2 + 9} + \frac{1}{x - 1} \] Find the domain.

Solution

\[ -x^2 + 9 \ge 0 \Rightarrow -3 \le x \le 3 \] \[ x \neq 1 \] Domain: \[ [-3,1) \cup (1,3] \]


Question 10

\[ f(x) = |x - 2| + 3 \] Find the range.

Solution

For \( x \) in \( \R \) \[ |x - 2| \ge 0 \Rightarrow |x - 2| + 3 \ge 3 \Rightarrow f(x) \ge 3 \] Range: \[ [3, \infty) \]


Question 11

\[ f(x) = -x^2 - 10 \] Find the range.

Solution

For \( x \) in \( \R \) \[ -x^2 \le 0 \Rightarrow -x^2 - 10 \le -10 \Rightarrow f(x) \le -10 \] Range: \[ (-\infty, -10] \]


Question 12

\[ h(x) = x^2 - 4x + 9 \] Find the range.

Solution

Complete the square and write \[ h(x) = (x - 2)^2 + 5 \] For \( x \) in \( \R \) \[ (x - 2)^2 \ge 0 \Rightarrow (x - 2)^2 + 5 \ge 5 \Rightarrow f(x) \ge 5 \] Range: \[ [5, \infty) \]


Question 13

\[ g(x) = \sqrt{x - 1}, \qquad h(x) = x^2 + 1 \] Find \((g \circ h)(x)\).

Solution

\[ (g \circ h)(x) = g(h(x)) = \sqrt{h(x) - 1} = \sqrt{x^2} = |x| \]


Question 14

Express the area \(A \) of a square as a function of its perimeter \( P \).

Solution

\[ P = 4x \Rightarrow x = \frac{P}{4} \] \[ A = x^2 = \frac{P^2}{16} \]

Exercises

  1. \(f(x) = |x - 6| + x^2 - 1\), find \(f(3)\)
  2. \(f(x) = ax + b\), find \(f(x+h)-f(x)\)
  3. Find the domain of \(\sqrt{-x^2 - x + 2}\)
  4. Find the range of \(-\sqrt{-x + 2} - 6\)
  5. \(f(x)=\sqrt{x},\, g(x)=x^2-2x+1\), find \((f\circ g)(x)\)

Answers

  1. 11
  2. \(ah\)
  3. \([-2,1]\)
  4. \((-\infty,-6]\)
  5. \(|x-1|\)