Functions: Practice Questions with Step-by-Step Solutions
Build confidence with function problems covering domain, evaluation, piecewise definitions, composite functions, absolute value, radical, and rational functions. Try each question first, then open the solution to compare your work with the detailed steps.
Core Concepts for Functions
- Function: each input \(x\) produces exactly one output \(y\).
- Domain: the set of all valid inputs.
- Piecewise function: defined by different formulas on different intervals.
- Composite function: \((f \circ g)(x) = f(g(x))\), evaluate the inner function first.
- Radical functions: the expression inside an even root must be \(\geq 0\).
- Rational functions: the denominator must not equal zero.
Practice Questions and Solutions
Question 1: Evaluating a piecewise function
Question: \(f\) is a function defined by
\[ f(x) = \begin{cases} 2x + 4 & x \leq 2 \\ 2x - 1 & x > 2 \end{cases} \]Find \(f(0)\), \(f(2)\), and \(f(4)\).
Solution:
Since \(0 \leq 2\), \(f(0)\) is found using the first formula \(f(x) = 2x + 4\):
\[ f(0) = 2(0) + 4 = 4 \]Since \(2 \leq 2\), \(f(2)\) is also found using the first formula:
\[ f(2) = 2(2) + 4 = 8 \]Since \(4 > 2\), \(f(4)\) is found using the second formula \(f(x) = 2x - 1\):
\[ f(4) = 2(4) - 1 = 7 \]Question 2: Identifying if an equation represents a function
Question: Is \(y\) a function of \(x\) in the equation \(-x^2 + y^2 - 2 = 0\)?
Solution:
Solve the equation for \(y\):
\[ y^2 = x^2 + 2 \] \[ y = \pm \sqrt{x^2 + 2} \]\(y\) is not a function of \(x\) because a single input \(x\) produces two distinct outputs \(y\).
Question 3: Finding the domain of a radical function
Question: Find the domain of the function \(f(x) = \sqrt{x^2 + 1} + 2\).
Solution:
The radicand must be non-negative. Since \(x^2 \geq 0\) for all real \(x\), we get:
\[ x^2 + 1 \geq 1 > 0 \]There are no restrictions on \(x\). The domain is the set of all real numbers, \((-\infty, \infty)\).
Question 4: Evaluating an absolute value function
Question: Let \(f(x) = |x - 1| + 3\). Find \(f(-9)\).
Solution:
\[ f(-9) = |-9 - 1| + 3 \] \[ f(-9) = |-10| + 3 = 10 + 3 = 13 \]Question 5: Evaluating a composite function
Question: Let \(f(x) = 2x + 2\) and \(g(x) = \sqrt{x + 3}\). Find \((f \circ g)(6)\).
Solution:
\[ (f \circ g)(6) = f(g(6)) \]First, evaluate \(g(6)\):
\[ g(6) = \sqrt{6 + 3} = \sqrt{9} = 3 \]Then apply \(f\) to \(3\):
\[ f(3) = 2(3) + 2 = 8 \]Thus, \((f \circ g)(6) = 8\).
Question 6: Finding roots of a rational function
Question: Find all real values of \(x\) such that \(f(x) = 0\), where \(f(x) = \dfrac{x^2 + 2x - 3}{x - 1}\).
Solution:
Set \(f(x) = 0\):
\[ \frac{x^2 + 2x - 3}{x - 1} = 0 \]A rational function is zero when its numerator is zero (and the denominator is nonzero). For \(x \neq 1\):
\[ x^2 + 2x - 3 = 0 \]Factor:
\[ (x - 1)(x + 3) = 0 \]This gives \(x = 1\) or \(x = -3\). Since \(x = 1\) makes the denominator zero, it is not valid. The only solution is \(x = -3\).
Question 7: Solving equations involving radical functions
Question: Find all real values of \(x\) such that \(f(x) = g(x)\), where \(f(x) = 3x + \sqrt{x}\) and \(g(x) = 2x + 6\).
Solution:
Start with \(f(x) = g(x)\):
\[ 3x + \sqrt{x} = 2x + 6 \]Isolate the radical:
\[ \sqrt{x} = -x + 6 \]Since \(\sqrt{x} \geq 0\), we also require \(-x + 6 \geq 0\), so \(x \leq 6\). Square both sides:
\[ x = (-x + 6)^2 \] \[ x = x^2 - 12x + 36 \] \[ x^2 - 13x + 36 = 0 \]Factor:
\[ (x - 4)(x - 9) = 0 \] \[ x = 4 \quad \text{or} \quad x = 9 \]Because we squared both sides, check for extraneous solutions:
- Check \(x = 4\):
LHS \(= 3(4) + \sqrt{4} = 12 + 2 = 14\)
RHS \(= 2(4) + 6 = 8 + 6 = 14\) ✔ Valid. - Check \(x = 9\):
LHS \(= 3(9) + \sqrt{9} = 27 + 3 = 30\)
RHS \(= 2(9) + 6 = 18 + 6 = 24\) ✘ Extraneous (also \(9 > 6\)).
Conclusion: \(x = 4\) is the only valid solution.
Question 8: Interpreting absolute value relations
Question: Is \(y\) a function of \(x\) in the equation \(|y + 1| = x^2 + 2\)?
Solution:
Clear the absolute value:
\[ y + 1 = \pm (x^2 + 2) \]This gives two possible values of \(y\) for a single \(x\):
\[ y = -1 - (x^2 + 2) \quad \text{or} \quad y = -1 + (x^2 + 2) \]Hence, \(y\) is not a function of \(x\).
Question 9: Domain of a function with absolute value and radical
Question: Find the domain of \(f(x) = \dfrac{1}{\sqrt{|x - 1|}}\).
Solution:
For \(f(x)\) to be real and defined, the denominator must be nonzero and the radicand must be positive. The expression \(|x - 1|\) equals zero only when \(x = 1\). So we must exclude \(x = 1\).
The domain is the set of all real numbers except \(x = 1\).
Key Takeaways
- Evaluate piecewise functions by first checking which interval the input belongs to.
- An equation defines \(y\) as a function of \(x\) only if each \(x\) gives exactly one \(y\).
- The domain of an even-root radical requires the radicand to be \(\geq 0\).
- For a composite function \((f \circ g)(x)\), always evaluate the inner function first.
- When solving radical equations, always check solutions — squaring can introduce extraneous roots.
- For rational functions, exclude any value that makes the denominator zero.