Precalculus Functions, Domain and Range Worksheet Solutions (PDF)

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This precalculus functions answers for Precalculus covers functions, domain and range (Common Core F-IF.A.1, F-IF.A.2, F-IF.B.5), with a free PDF to print.

Use this key to check each worked solution. Every answer shows the substitution or inequality used, and the final answer is in bold.

Precalculus · Common Core State Standards: F-IF.A.1, F-IF.A.2, F-IF.B.5

Precalculus functions answers: exercises 1 to 10

Exercise 1 – Evaluate a quadratic

Substitute \(x=3\): \(2(9)-5(3)+1=18-15+1\).

\(f(3)=4\)

Exercise 2 – Evaluate a cubic

\(g(-2)=(-2)^{3}-4(-2)=-8+8\).

\(g(-2)=0\)

Exercise 3 – Is it a function?

The input \(2\) appears twice with different outputs, \(5\) and \(7\).

Not a function

Exercise 4 – Domain of a line

A linear function has no division and no roots, so every real number is allowed.

\((-\infty,\infty)\)

Exercise 5 – Domain with a denominator

Set \(x-4=0\), so \(x=4\) is excluded.

\((-\infty,4)\cup(4,\infty)\)

Exercise 6 – Domain with a root

Require \(2x+6\ge 0\), so \(2x\ge -6\) and \(x\ge -3\).

\([-3,\infty)\)

Exercise 7 – Root and denominator together

The root needs \(x+5\ge 0\), so \(x\ge -5\). The denominator excludes \(x=2\), which lies inside that interval.

\([-5,2)\cup(2,\infty)\)

Exercise 8 – Range of a shifted parabola

Since \(x^{2}\ge 0\), we have \(f(x)\ge 3\), and \(3\) is reached at \(x=0\).

\([3,\infty)\)

Exercise 9 – Read a graph

-2-11234-112345xy(-1, 0)(3, 0)(1, 4)
Domain runs from x = -1 to x = 3; the highest point is (1, 4).

The graph runs from \(x=-1\) to \(x=3\) with closed endpoints, so the domain is \([-1,3]\). The lowest \(y\) is \(0\) and the highest is \(4\), so the range is \([0,4]\). Using \(f(x)=-(x-1)^{2}+4\): \(f(2)=-1+4=3\).

Domain \([-1,3]\); range \([0,4]\); \(f(2)=3\)

Exercise 10 – Piecewise values

\(-2< 1\), so \(f(-2)=-2+4=2\). \(1\ge 1\), so \(f(1)=1^{2}=1\). \(3\ge 1\), so \(f(3)=9\).

\(2,\ 1,\ 9\)

Exercises 11 and up

Exercise 11 – Vertical line test

-4-3-2-11234-4-3-2-11234xy
The vertical line x = 1.5 meets the circle at two points.

When \(x=1.5\), \(y=\pm\sqrt{9-2.25}=\pm\sqrt{6.75}\approx\pm 2.6\). One input gives two outputs.

No, it fails the vertical line test

Exercise 12 – Domain of a model

Both the width \(w\) and the length \(20-w\) must be positive, so \(w>0\) and \(w< 20\).

\((0,20)\)

Exercise 13 – Reciprocal of a root

The radicand must be positive (it cannot be 0, since that would put 0 in the denominator): \(x-3>0\).

\((3,\infty)\)

Exercise 14 – Solve f(x) = value

\(4x-1=11\), so \(4x=12\).

\(x=3\)

Exercise 15 – Input for a quadratic output

\(x^{2}-2x-8=0\), so \((x-4)(x+2)=0\).

\(x=4\) or \(x=-2\)

Exercise 16 – Difference quotient

\(f(2+h)=4+4h+h^{2}+1=5+4h+h^{2}\) and \(f(2)=5\). The numerator is \(4h+h^{2}\); divide by \(h\).

\(4+h\)

Exercise 17 – Taxi fare

\(C(6.4)=4+2.5(6.4)=4+16=20\). For \(\$34\): \(4+2.5m=34\), so \(2.5m=30\) and \(m=12\).

\(\$20.00\); \(12\) miles

Exercise 18 – Height of a ball

1234520406080t (s)h (ft)(2, 69)
The ball peaks at (2, 69).

\(h(0)=5\). \(h(1)=-16+64+5=53\). \(h(2)=-64+128+5=69\). The vertex is at \(t=-\frac{64}{2(-16)}=2\), so the maximum is \(69\) ft. The ball ends at height \(0\).

\(5,\ 53,\ 69\); range \([0,69]\)

Exercise 19 – Data plan

For \(3\) and \(5\) GB the cost is the flat \(\$25\). For \(9.5\) GB: \(25+8(9.5-5)=25+36=61\). The cost is never below \(25\) and grows without bound.

\(\$25,\ \$25,\ \$61\); range \([25,\infty)\)

Exercise 20 – Semicircle function

Require \(9-x^{2}\ge 0\), so \(-3\le x\le 3\). The output is at most \(f(0)=3\) and at least \(f(\pm 3)=0\).

Domain \([-3,3]\); range \([0,3]\)

Key facts behind this precalculus functions answers

Key facts
  • A relation is a function only if each input has exactly one output (vertical line test).
  • Domain: allowed inputs. Range: outputs that actually occur.
  • Exclude values that make a denominator 0 or an even-root radicand negative.
  • Interval notation uses brackets for included endpoints and parentheses for excluded ones.
  • A piecewise function uses a different rule on each part of its domain.

How to use this precalculus functions answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

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