Functions: Domain, Range, Composition and Inverses: Precalculus Math Worksheet Answer Key – PDF Download

Precalculus Functions Review Worksheet Answer Key: problems 1–5

Problem 1: Evaluate function notation

1. \(f(3)=3(9)-6+1=22\), so \(22\)
2. \(f(-2)=3(4)+4+1=17\), so \(17\)
3. \(f(a+1)=3(a^{2}+2a+1)-2a-2+1=3a^{2}+4a+2\), so \(3a^{2}+4a+2\)

Problem 2: Domain with square roots

1. \(2x-6 \ge 0\) gives \(2x \ge 6\) and \(x \ge 3\). Domain: \([3,\infty)\)
2. \(10-5x \ge 0\) gives \(-5x \ge -10\); dividing by \(-5\) reverses the inequality, so \(x \le 2\). Domain: \((-\infty,2]\)

Problem 3: Domain with denominators

1. \(x^{2}-9=0\) when \(x=3\) or \(x=-3\), so those two values are excluded. Domain: \((-\infty,-3)\) ∪ \((-3,3)\) ∪ \((3,\infty)\)
2. \(x^{2}+4 \ge 4\) is never zero, so every real number works. Domain: \((-\infty,\infty)\)

Problem 4: Domain with several restrictions

1. \(x+4 \ge 0\) gives \(x \ge -4\), and \(x \neq 2\) removes a zero denominator. Domain: \([-4,2)\) ∪ \((2,\infty)\)
2. The logarithm needs \(5-x > 0\), so \(x < 5\). Domain: \((-\infty,5)\)

Problem 5: Domain and range from a graph

1. The graph runs from \(x=-1\) (included) to \(x=3\) (excluded). Domain: \([-1,3)\)
2. The lowest height is \(0\) (at the closed point) and the peak is \(4\). Range: \([0,4]\)
3. The graph is \(-(x-1)^{2}+4\). Solve \(-(x-1)^{2}+4=3\): \((x-1)^{2}=1\), so \(x=0\) or \(x=2\). Answer: \(x=0\) and \(x=2\)

Precalculus Functions Review Worksheet Answer Key: problems 6–10

Problem 6: Composition from tables

1. \(g(1)=3\) and \(f(3)=1\), so \(1\)
2. \(f(3)=1\) and \(g(1)=3\), so \(3\)
3. \(f(1)=2\) and \(f(2)=4\), so \(4\)
4. \(g(4)=2\) and \(g(2)=1\), so \(1\)

Problem 7: Composition of polynomials

1. \(f(g(x))=(3x-2)^{2}+1=9x^{2}-12x+4+1\), so \(9x^{2}-12x+5\)
2. \(g(f(x))=3(x^{2}+1)-2\), so \(3x^{2}+1\)
3. \(g(2)=4\) and \(f(4)=17\). Check: \(9(4)-24+5=17\). Answer: \(17\)

Problem 8: Composition from a mapping diagram

1. Follow each arrow twice: \(1 \to 3 \to 9\), \(2 \to 5 \to 25\), \(3 \to 7 \to 49\). Answers: \(9, 25, 49\)
2. The outputs of \(g\) rise by 2 each step from \(3\), so \(g(x)=2x+1\). The outputs of \(f\) are squares, so \(f(x)=x^{2}\). Then \(f(g(x))=(2x+1)^{2}\); check \((2\cdot 3+1)^{2}=49\).

Problem 9: Domain of a composition (roots)

1. \(f(g(x))=\sqrt{x-5}\) needs \(x-5 \ge 0\). Domain: \([5,\infty)\)
2. \(g(f(x))=\sqrt{x}-5\) needs \(x \ge 0\). Domain: \([0,\infty)\)

Problem 10: Domain of a composition (rational)

1. \(f(g(x))=\frac{1}{x+2}\), and \(x+2 \neq 0\). Domain: \((-\infty,-2)\) ∪ \((-2,\infty)\)
2. \(g(f(x))=\frac{1}{x}+2=\frac{2x+1}{x}\), and \(x \neq 0\). Domain: \((-\infty,0)\) ∪ \((0,\infty)\)

Precalculus Functions Review Worksheet Answer Key: problems 11–15

Problem 11: Decompose a function

1. The inside is \(2x+5\): \(g(x)=2x+5\), \(f(x)=x^{4}\)
2. The inside is \(x-3\): \(g(x)=x-3\), \(f(x)=\frac{1}{x^{2}}\)
3. The inside is \(x^{2}+9\): \(g(x)=x^{2}+9\), \(f(x)=\sqrt{x}\). (Other correct splits exist.)

Problem 12: Inverse of a linear function

1. Swap: \(x=5y-3\), so \(y=\frac{x+3}{5}\). Answer: \(f^{-1}(x)=\frac{x+3}{5}\)
2. Swap: \(x=-\frac{2}{3}y+4\), so \(-\frac{2}{3}y=x-4\) and \(y=-\frac{3}{2}(x-4)=6-\frac{3}{2}x\). Answer: \(g^{-1}(x)=6-\frac{3}{2}x\). Check: \(g(6)=0\) and \(g^{-1}(0)=6\).

Problem 13: Inverse of a rational function

Swap: \(x=\frac{2y+1}{y-3}\). Then \(xy-3x=2y+1\), so \(y(x-2)=3x+1\).

Answer: \(f^{-1}(x)=\frac{3x+1}{x-2}\). Check: \(f(4)=9\) and \(f^{-1}(9)=\frac{28}{7}=4\).

The domain of \(f\) excludes \(x=3\); the domain of \(f^{-1}\) excludes \(x=2\), which is the range of \(f\). Domain of \(f\): \(x \neq 3\); range: \(y \neq 2\).

Problem 14: Verify inverse functions

1. \(f(g(x))=(\sqrt[3]{x-4})^{3}+4=x\) and \(g(f(x))=\sqrt[3]{x^{3}+4-4}=x\). Yes, inverses.
2. \(f(g(x))=\frac{1}{2}(2x+3)-3=x-\frac{3}{2}\), which is not \(x\). Not inverses; the true inverse is \(2x+6\).

Problem 15: Inverse values from a graph

1. \((2,1)\) is on the graph, so \(f(2)=1\)
2. The output \(3\) comes from \(x=4\), so \(f^{-1}(3)=4\)
3. The output \(-1\) comes from \(x=0\), so \(f^{-1}(-1)=0\)
4. The output \(7\) comes from \(x=6\), so \(f^{-1}(7)=6\)

Precalculus Functions Review Worksheet Answer Key: problems 16–20

Problem 16: Inverse values from tables

1. The output \(0\) occurs at \(x=2\) and the output \(-2\) at \(x=3\). Answers: \(f^{-1}(0)=2\) and \(f^{-1}(-2)=3\)
2. \(f(1)=3\) and \(f^{-1}(3)=1\), so \(1\)
3. \(g(1)=4\) and \(g(3)=4\): two inputs share one output, so reversing the table would give two answers for \(4\). \(g\) is not one-to-one.

Problem 17: Horizontal line test

Graph A rises everywhere, so every horizontal line meets it once: invertible. A horizontal line at \(y=1\) meets the parabola in graph B twice, and one at \(y=0.3\) meets graph C three times . Only graph A has an inverse function.

Problem 18: Restrict a domain: vertex form

1. The graph fails the horizontal line test: \(f(1)=4\) and \(f(5)=4\).
2. Swap: \(x=(y-3)^{2}\). On the restricted domain \(y \ge 3\), so \(y-3=\sqrt{x}\). Answer: \(f^{-1}(x)=3+\sqrt{x}\), domain \([0,\infty)\), range \([3,\infty)\)
3. \(f^{-1}(16)=3+\sqrt{16}=7\). Check: \(f(7)=16\). Answer: \(7\)

Problem 19: Restrict a domain: complete the square

Complete the square: \(f(x)=(x+2)^{2}-4\). Swap: \(x=(y+2)^{2}-4\), so \((y+2)^{2}=x+4\). Since \(y \ge -2\), \(y+2=\sqrt{x+4}\).

Answer: \(f^{-1}(x)=-2+\sqrt{x+4}\), domain \([-4,\infty)\) (the range of \(f\)), range \([-2,\infty)\). Check: \(f(1)=5\) and \(f^{-1}(5)=-2+3=1\).

Problem 20: Temperature conversion

1. Swap and solve: \(C=\frac{5}{9}(F-32)\) gives \(F-32=\frac{9}{5}C\). Answer: \(F(C)=\frac{9}{5}C+32\)
2. \(F(37)=\frac{9}{5}(37)+32=66.6+32=98.6\). Answer: \(98.6^{\circ}\text{F}\)
3. \(C(F(C))=\frac{5}{9}(\frac{9}{5}C+32-32)=\frac{5}{9}\cdot\frac{9}{5}C=C\). Verified.

Precalculus Functions Review Worksheet Answer Key: problems 21–24

Problem 21: Coupon and discount order

1. \(d(60)=48\) and \(c(48)=43\); \(c(60)=55\) and \(d(55)=44\). Answers: $43 and $44
2. \(c(d(p))=0.8p-5\) and \(d(c(p))=0.8(p-5)=0.8p-4\). Answers: \(0.8p-5\) and \(0.8p-4\)
3. Taking the discount first and the coupon second gives the lower price. The difference is \((0.8p-4)-(0.8p-5)=1\), so it saves $1 at every price.

Problem 22: Taxi fare inverse

1. Solve \(F=2.50+1.80m\) for \(m\): \(m=\frac{F-2.50}{1.80}\). It converts a fare into the number of miles ridden.
2. \(m(20.50)=\frac{18.00}{1.80}=10\). Answer: 10 miles
3. \(m(11.50)=\frac{9.00}{1.80}=5\). Answer: 5 miles

Problem 23: Growing ripple

1. \(A(r(t))=\pi(3t)^{2}\). Answer: \(9\pi t^{2}\)
2. \(9\pi(4)^{2}=144\pi\approx 452.4\). Answer: 452.4 square feet
3. \(9\pi t^{2}=81\pi\) gives \(t^{2}=9\), and \(t > 0\), so \(t=3\) seconds

Problem 24: Inverse of a composition

1. \(f(g(x))=2(x-4)+1=2x-7\). Swap: \(x=2y-7\), so the inverse is \(\frac{x+7}{2}\)
2. \(f^{-1}(x)=\frac{x-1}{2}\) and \(g^{-1}(x)=x+4\). Then \(g^{-1}(f^{-1}(x))=\frac{x-1}{2}+4=\frac{x+7}{2}\). Answer: \(\frac{x+7}{2}\)
3. They match. The inverse of \(f(g(x))\) undoes \(f\) first and then \(g\): \(g^{-1}(f^{-1}(x))\).

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