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

Precalculus Functions Review Test Answer Key: questions 1–5

Question 1: Domain in interval notation (2 points)

1. \(3x-12 \ge 0\) gives \(x \ge 4\): \([4,\infty)\)
2. \(x^{2}-x-6=(x-3)(x+2)\) is zero at \(3\) and \(-2\): \((-\infty,-2)\) ∪ \((-2,3)\) ∪ \((3,\infty)\)

Scoring: 1 point for each correct domain; deduct half of a point for a wrong bracket type.

Question 2: Domain and range from a graph (2 points)

1. The graph begins at \(x=-1\) and continues right forever: \([-1,\infty)\)
2. The graph begins at height \(3\) and goes down forever: \((-\infty,3]\)

Scoring: 1 point for each interval written with correct endpoints and brackets.

Question 3: Composition of functions (2 points)

1. \(f(g(x))=(x+4)^{2}-3(x+4)=x^{2}+8x+16-3x-12\): \(x^{2}+5x+4\)
2. \(f(2)=4-6=-2\) and \(g(-2)=2\): \(2\)

Scoring: Part (a): 1 point for the correct simplified polynomial. Part (b): 1 point for the correct value.

Question 4: Decompose and find a domain (2 points)

1. \(g(x)=x^{2}-16\) and \(f(x)=\sqrt{x}\)
2. \(x^{2}-16 \ge 0\) means \(x \le -4\) or \(x \ge 4\): \((-\infty,-4]\) ∪ \([4,\infty)\)

Scoring: 1 point for a valid decomposition; 1 point for the correct domain (half credit if only one piece is given).

Question 5: Find and verify an inverse (3 points)

1. Swap: \(x=\frac{3y-2}{y+1}\), so \(xy+x=3y-2\) and \(y(x-3)=-x-2\). \(f^{-1}(x)=\frac{x+2}{3-x}\)
2. The numerator is \(\frac{3x-2+2x+2}{x+1}=\frac{5x}{x+1}\) and the denominator is \(\frac{3x+3-3x+2}{x+1}=\frac{5}{x+1}\). Dividing gives \(x\).

Scoring: Part (a): 1 point for swapping and clearing the fraction, 1 point for the correct inverse. Part (b): 1 point for a complete verification.

Precalculus Functions Review Test Answer Key: questions 6–9

Question 6: Inverse values from a graph (3 points)

1. The output \(2\) occurs at \(x=0\): \(f^{-1}(2)=0\)
2. The output \(6\) occurs at \(x=4\): \(f^{-1}(6)=4\)
3. Domain of \(f^{-1}\) is the range of \(f\): \([-3,6]\); range of \(f^{-1}\) is the domain of \(f\): \([-4,4]\)

Scoring: 1 point for each part; in part (c) both intervals are needed.

Question 7: Restrict a domain (3 points)

1. A horizontal line such as \(y=0\) meets the graph at \(x=1\) and \(x=5\), so \(f\) is not one-to-one.
2. Complete the square: \(f(x)=(x-3)^{2}-4\). Swap: \(x=(y-3)^{2}-4\), so \((y-3)^{2}=x+4\) and, since \(y \ge 3\), \(f^{-1}(x)=3+\sqrt{x+4}\), with domain \([-4,\infty)\).

Scoring: Part (a): 1 point for a valid reason (two inputs, one output). Part (b): 1 point for the inverse formula, 1 point for the domain.

Question 8: Temperature inverse (2 points)

1. \(F(C)=\frac{9}{5}C+32\)
2. \(F(-15)=\frac{9}{5}(-15)+32=-27+32=5\): \(5^{\circ}\text{F}\)

Scoring: 1 point for each correct part.

Question 9: Find an unknown function (1 point)

Since \(6x-7=2(3x-5)+3\), \(f(x)=2x+3\). Check: \(f(g(x))=2(3x-5)+3=6x-7\).

Scoring: 1 point for the correct function.

Scoring summary

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