ACT Math study guide · Chapter 12
Function Notation and Composition
Function notation f(x) names a rule: whatever value replaces x on the right side of the equation gets substituted everywhere x appears. Evaluating a function at a specific input is direct substitution. When the input itself is an expression, such as f(a+2), every instance of x in the rule gets replaced by that entire expression, not just the letter.
Composition means evaluating one function, then feeding that result into another. f(g(x)) always evaluates the inner function first: find g(x), then substitute that number into f. The order matters. f(g(x)) and g(f(x)) are generally different functions, since composition is not commutative.
The domain of a function is every input value that produces a valid output. Several conditions can restrict a domain on this test: a denominator cannot equal zero, the expression under an even root cannot be negative, and the argument of a logarithm must be positive. Each condition is solved as an inequality or exclusion, not evaluated as if solving for a specific value. When a function combines more than one of these restrictions, every condition must hold at once, so the domain is the intersection of all of them, not just the easiest one to spot.
The range is the set of output values a function can actually produce, the counterpart to domain: domain answers which x-values are allowed, range answers which y-values result. From a graph, the range is the set of y-values actually attained by points on the graph. Use open and closed endpoints, holes, and arrows to decide which boundary values are excluded and whether the range continues without bound. From a table, the range is simply the set of output values shown. For a transformed basic function, track what each transformation does to the original function's range: a vertical shift moves the whole range up or down, and a reflection over the x-axis flips its sign.
Where this appears as a model: a described real-world rule. If temperature drops 3°F for every 1,000 feet of altitude starting from 68°F, the function F(h)=68−3h gives F(4)=56°F at 4,000 feet.
▸ Composition, step by step
Worked Example Let f(x) = x + 5 and g(x) = x². Find f(g(3)).
Evaluate the inner function first.
g(3) = 3² = 9
f(9) = 9 + 5 = 14
▸ Finding a function's domain
Worked Example Find the domain of h(x) = √(x − 3) ⁄ (x − 8).
Two conditions apply at once: the radical and the denominator.
Radical: x − 3 ≥ 0 → x ≥ 3
Denominator: x − 8 ≠ 0 → x ≠ 8
Domain: x ≥ 3 and x ≠ 8
▸ Practice
1. Let f(x) = 3x − 4. Find f(5).
A) 7 B) 19 C) 11 D) −7
2. Let g(x) = x² + 2. Find g(−4).
F) −14 G) 16 H) 18 J) 14
3. Let f(x) = 2x + 1 and g(x) = x − 3. Find f(g(4)).
A) 9 B) 3 C) 6 D) 1
4. Let f(x) = x² − 1 and g(x) = x + 2. Find g(f(2)).
F) 15 G) 5 H) 3 J) 1
5. Find the domain of h(x) = √(x + 2) ⁄ (x − 5).
A) x ≥ −2 and x ≠ 5 B) x > −2 and x ≠ 5 C) x ≥ 2 and x ≠ 5 D) x ≥ −2, no other restriction
6. Let f(x) = x² and g(x) = 2x − 1. For what value(s) of x does f(g(x)) = 9?
F) x = 2 or x = −1 G) x = 2 only H) x = −1 only J) x = 1 or x = −2
7. Let f(x) = 5x − 2. Find f(−3).
A) −17 B) −13 C) 17 D) 13
8. Let f(x) = x² + 1 and g(x) = 2x − 3. Find f(g(1)).
F) −1 G) 5 H) 2 J) −4
9. Find the domain of h(x) = 1 ⁄ (x − 6).
A) x ≠ 6 B) x ≠ 0 C) x ≥ 6 D) x ≤ 6
10. Let f(x) = 3x + 4 and g(x) = x − 1. Find g(f(2)).
F) 10 G) 7 H) 11 J) 9
11. Find the domain of f(x) = √(4 − x).
A) x ≥ 4 B) x ≤ 4 C) x ≤ −4 D) x ≥ −4
12. Let f(x) = 2x − 1 and g(x) = x² + 2. Find f(g(−2)).
F) 6 G) 9 H) 13 J) 11
13. Let f(x) = 2x and g(x) = x² − 5. Find f(g(3)).
A) 4 B) 1 C) 8 D) 13
14. If f(3) = 7 and g(x) = 2x − 1, find g(f(3)).
F) 13 G) 7 H) 6 J) 5
15. Let f(x) = 4x − 7. Find f(3).
A) 12 B) −5 C) 19 D) 5
16. Let f(x) = x² − 2 and g(x) = 3x + 1. Find f(g(0)).
F) 1 G) −1 H) −2 J) 7
17. Find the domain of h(x) = √(5x − 10).
A) x ≤ 2 B) x ≥ −2 C) x ≤ −2 D) x ≥ 2
18. Let f(x) = 2x + 3 and g(x) = x − 4. Find g(f(1)).
F) 1 G) 5 H) 9 J) −3
19. Let f(x) = x² and g(x) = 2x − 5. For what value(s) of x does f(g(x)) = 1?
A) x = 3 only B) x = 3 or x = 2 C) x = 2 only D) no solution
20. Find the domain of h(x) = 1 ⁄ (x + 9).
F) x ≠ 9 G) x ≠ −9 H) x ≥ −9 J) x ≤ −9
▸ Answers and Explanations
1: C. 3(5) − 4 = 15 − 4 = 11. Choice B adds 4 instead of subtracting it, computing 3(5)+4=19.
2: H. (−4)² + 2 = 16 + 2 = 18. Choice F treats (−4)² as −16 instead of +16, a sign error when squaring a negative number.
3: B. g(4) = 4−3 = 1, then f(1) = 2(1)+1 = 3. Choice A evaluates f at 4 directly instead of first evaluating g(4).
4: G. f(2) = 2²−1 = 3, then g(3) = 3+2 = 5. Choice F reverses the order of composition, computing f(g(2)) instead of g(f(2)).
5: A. The radical requires x+2≥0 (x≥−2); the denominator requires x≠5. Choice B uses a strict inequality for the radical condition instead of ≥, excluding the boundary value where the expression equals zero.
6: F. (2x−1)²=9 gives 2x−1=3 or 2x−1=−3, so x=2 or x=−1. Choice G reports only one of the two solutions from the squared equation.
7: A. 5(−3)−2=−15−2=−17. Choice B adds 2 instead of subtracting it, computing 5(−3)+2 = −13.
8: H. g(1)=2(1)−3=−1; f(−1)=(−1)²+1=2. Choice F reports the inner value g(1) instead of completing the composition by evaluating f at that value.
9: A. The denominator cannot equal zero: x≠6. Choice B uses 0 instead of the actual value that makes the denominator zero.
10: J. f(2)=3(2)+4=10; g(10)=10−1=9. Choice F reports the inner value f(2) instead of completing the composition by evaluating g at that value.
11: B. The radical requires 4−x≥0, so x≤4. Choice A reverses the inequality direction required for the expression under the radical to be nonnegative.
12: J. g(−2)=(−2)²+2=6; f(6)=2(6)−1=11. Choice F reports the inner value g(−2) instead of completing the composition by evaluating f at that value.
13: C. g(3)=3²−5=4; f(4)=2(4)=8. Choice A reports the inner value g(3) instead of completing the composition by evaluating f at that value.
14: F. g(7)=2(7)−1=13. Choice G reports the given value f(3) again instead of applying g to that value.
15: D. 4(3)−7=12−7=5. Choice A reports 4(3) directly, forgetting to subtract 7.
16: G. g(0)=3(0)+1=1; f(1)=1²−2=−1. Choice F reports g(0) squared without subtracting 2, forgetting the second part of f's rule.
17: D. The radical requires 5x−10≥0, so x≥2. Choice A reverses the inequality direction required for the expression under the radical to be nonnegative.
18: F. f(1)=2(1)+3=5; g(5)=5−4=1. Choice G reports the inner value f(1) instead of completing the composition by evaluating g at that value.
19: B. (2x−5)²=1, so 2x−5=±1; 2x−5=1 gives x=3; 2x−5=−1 gives x=2. Choice A reports only one of the two solutions from the squared equation.
20: G. The denominator cannot equal zero: x≠−9. Choice F drops the negative sign on the excluded value.
▸ Domain and range across representations
Domain and range questions do not always arrive as a bare algebraic expression. A graph shows domain along the x-axis and range along the y-axis directly. A real-world context can restrict a domain further than the algebra alone would suggest. A composition can hide a domain restriction on the inner function's output, which then has to satisfy the outer function's own domain.
1. A graph exists only for x-values from −3 to 5, with y-values ranging from −2 to 6 over that interval. State the domain and range shown by the graph.
A) Domain: −2≤x≤6; Range: −3≤y≤5 B) Domain: −3≤x≤5; Range: −2≤y≤6 C) Domain: all reals; Range: −2≤y≤6 D) Domain: −3≤x≤5; Range: all reals
2. Find the domain of f(x) = log(x+2) ÷ (x−5).
F) x>−2 G) x≠ 5 H) x>−2 and x≠5 J) x≥−2 and x≠5
3. A ball's height in feet after t seconds is modeled by h(t) = −16t²+64t. The algebraic domain of this quadratic is all real numbers. What is the domain in the context of the ball's actual flight?
A) All real numbers B) t≥0 C) 0≤t≤4 D) t≥4
4. Let f(x)=√x and g(x)=x−10. Find the domain of f(g(x)).
F) x≥0 G) x≤10 H) x≥10 J) All real numbers
▸ Answers and Explanations
1: B. Domain is read along the x-axis (−3 to 5); range is read along the y-axis (−2 to 6). Choice A swaps the domain and range intervals.
2: H. The logarithm requires x+2>0, so x>−2; the denominator requires x−5≠0, so x≠5. Both restrictions must hold together. Choice F reports only the logarithm's restriction, dropping the denominator exclusion.
3: C. The ball's flight begins at t=0 and ends when it lands: −16t(t−4)=0, giving t=0 or t=4. Only the time between launch and landing makes sense in context, so the domain is 0≤t≤4. Choice A applies the unrestricted algebraic domain, ignoring that the model only describes the ball while it's actually in the air.
4: H. The inner function g(x)=x−10 must produce a value that satisfies f's domain, x≥0, so x−10≥0, giving x≥10. Choice F reports g's output restriction with the wrong shift, as if g(x)=x instead of x−10.