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Quizzes > High School Quizzes > Mathematics

Function Operations Quick Check Quiz

Master function operations through targeted quick practice

Difficulty: Moderate
Grade: Grade 9
Study OutcomesCheat Sheet
Paper art promoting the Function Ops Blitz practice quiz for high school algebra students.

Given f(x) = 2x + 3, what is f(4)?
11
10
9
8
To evaluate f(4), substitute 4 in place of x: 2*4 + 3 = 11. This is a straightforward linear function evaluation.
What is the domain of the function f(x) = √(x - 1)?
x ≥ 1
x > 1
x ≤ 1
x < 1
Since the square root requires a non-negative input, the expression inside the root must satisfy x - 1 ≥ 0. Thus, the domain is x ≥ 1.
For the function g(x) = x², what is g(3)?
9
6
3
8
Substituting 3 in for x yields 3², which equals 9. This is a simple power function evaluation.
What is the result of adding the functions f(x) = x + 2 and g(x) = 3x?
4x + 2
x + 5
3x + 2
2x + 2
Adding the two functions gives: (x + 2) + 3x = 4x + 2. The like terms combine directly.
Determine the output of the function h(x) = 5 when x = -10.
5
0
-10
5x
h(x) is a constant function, meaning its output does not change regardless of the input. Therefore, h(-10) is 5.
Find the value of (f ∘ g)(2) if f(x) = x + 4 and g(x) = 3x - 1.
9
10
11
8
First compute g(2): 3(2) - 1 = 5. Then f(5) = 5 + 4 = 9. This demonstrates function composition in a linear context.
Given f(x) = 2x - 3 and g(x) = x², compute (g ∘ f)(3).
9
6
3
12
First, find f(3): 2(3) - 3 = 3. Then, g(3) = 3² = 9. This is a clear example of composing a linear function with a quadratic function.
If f(x) = √x and g(x) = x - 1, what is (f ∘ g)(5)?
2
4
√5
3
Compute g(5) which is 5 - 1 = 4. Then apply f to get √4 = 2. The process reinforces the steps in function composition.
For the function f(x) = 1/(x - 2), what is its domain?
All real numbers except 2
All positive real numbers
x ≠ -2
All real numbers
Since division by zero is undefined, the denominator x - 2 cannot equal zero. Thus, x cannot be 2.
Determine (f + g)(x) if f(x) = 3x and g(x) = x².
x² + 3x
3x² + x
3x² + 3x
x² - 3x
The sum of the functions is computed by adding corresponding expressions: 3x + x², which is typically written as x² + 3x. This showcases function addition.
Find (f - g)(x) if f(x) = 4x + 5 and g(x) = 2x - 3.
2x + 8
2x + 2
6x + 2
2x - 8
Subtracting g(x) from f(x) leads to: (4x + 5) - (2x - 3) = 2x + 8. This question emphasizes careful handling of subtraction and like terms.
If f(x) = x³ and g(x) = 2x, compute (f ∘ g)(2).
64
8
16
32
First compute g(2) = 2(2) = 4, then substitute into f to get 4³ = 64. This problem reinforces the concept of function composition with polynomial functions.
Evaluate the composition (g ∘ f)(-1) if f(x) = x + 2 and g(x) = x².
1
0
-1
4
Start by evaluating f(-1) to get (-1) + 2 = 1, then square the result using g: 1² = 1. The process underlines the sequential nature of composition.
If f(x) = |x| and g(x) = x - 3, what is f(g(5))?
2
8
-2
5
Calculate g(5): 5 - 3 = 2, then apply the absolute value function f to obtain |2| = 2. This reinforces the use of absolute value in function operations.
Given the function f(x) = 1/(2 - x), identify the value that is not in the domain of f.
2
0
-2
1
The denominator 2 - x equals zero when x = 2, making the function undefined at that point. Therefore, x = 2 is excluded from the domain.
Let f(x) = 2x + 1 and g(x) = x². For which values of x does (f ∘ g)(x) equal f(x)?
x = 0 or 1
x = 1
x = 0
x = -1 or 1
Substituting into the composition, (f ∘ g)(x) becomes 2x² + 1. Setting this equal to f(x), which is 2x + 1, yields 2x² = 2x and then x² = x. This factors to x(x - 1) = 0, so x = 0 or x = 1.
Given f(x) = (x - 3)/(x + 2) and g(x) = (2x + 4)/(x - 1), determine the simplified form of (f ∘ g)(x) (excluding domain restrictions).
(7 - x) / [2(2x + 1)]
(x - 7) / [2(2x + 1)]
(7 + x) / [2(2x + 1)]
(x - 7) / [2(x + 1)]
By substituting g(x) into f(x) and simplifying the resulting complex fraction, the expression reduces to (7 - x) divided by 2(2x + 1). The cancellation of the common factor (x - 1) is crucial in this derivation.
If f(x) = log(x) (base 10) and g(x) = 10^x, find (f ∘ g)(x).
x
10x
log(x)
10^x
The logarithm and the exponential function with a matching base are inverses. Therefore, composing them yields x.
Solve for x in (f ∘ g)(x) = 5, where f(x) = x + 2 and g(x) = 3x.
1
5
3
7/3
First, compute the composition: f(g(x)) = 3x + 2. Setting 3x + 2 = 5 and solving for x gives x = 1. This is a typical example of solving a linear equation.
Let f(x) = √(x + 4) and g(x) = x² - 4. Determine the composite function (f ∘ g)(x) and identify its domain restriction.
|x|, all real numbers
x², all real numbers
√(x) with x ≥ 0
x with x ≥ 0
The composition f(g(x)) simplifies to √((x² - 4) + 4) = √(x²) which equals |x|. Since x² is always non-negative, there is no additional restriction on x.
0
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Study Outcomes

  1. Analyze and simplify composite functions.
  2. Evaluate function operations with accuracy.
  3. Interpret the effects of function transformations.
  4. Apply properties of operations to solve function-based problems.
  5. Verify results through back-substitution and alternative methods.

Function Operations Quick Check Cheat Sheet

  1. Master function operations - Adding, subtracting, multiplying, and dividing functions is like stacking blocks to create amazing structures. For example, if f(x)=x² and g(x)=x+1, then (f+g)(x)=x²+x+1. Once you've got these moves down, you can tackle any function mash‑up confidently. Pearson: Function Operations
  2. Determine domains of combined functions - The domain is your function's legal playground; when you divide functions, make sure you never set the denominator to zero or you'll break the rules. For instance, if g(x)=x - 2, (f/g)(x) is undefined at x=2. Keeping track of these spots avoids nasty surprises. Pearson: Domain Rules
  3. Practice function composition - Composition is like nesting dolls: you fit one function inside another. With f(g(x)), you replace x in f with the entire g(x). For example, if f(x)=2x and g(x)=x+3, f(g(x))=2(x+3)=2x+6, unlocking powerful ways to build complex formulas. Pearson: Composition Tips
  4. Explore inverse functions - An inverse function "undoes" its partner: f❻¹(f(x))=x. To find one, swap x and y in y=f(x) then solve for y. For f(x)=2x+3, you get f❻¹(x)=(x - 3)/2, which is like using a secret code breaker to reverse calculations. SparkNotes: Inverse Functions
  5. Visualize transformations - Shifts, reflections, stretches, and shrinks turn graphs into shape‑shifting stars. For example, f(x - 2) slides f(x) two units right, and f( - x) flips it over the y‑axis. Spotting these changes helps you sketch and interpret graphs like a pro. SparkNotes: Transformations
  6. Apply the FOIL method - When two binomials collide, FOIL (First, Outer, Inner, Last) saves the day. For instance, (x+2)(x - 3)=x² - 3x+2x - 6=x² - x - 6. Mastering FOIL speeds up multiplying functions and keeps your algebra slick. Pearson: FOIL Guide
  7. Spot even and odd functions - Spotting symmetry is fun: even functions satisfy f( - x)=f(x) and mirror over the y‑axis, while odd functions satisfy f( - x)= - f(x) and rotate 180°. Knowing these properties makes graph analysis a breeze. SparkNotes: Symmetry Rules
  8. Nail function notation - Function notation f(x) is your shorthand for "plug x in here." If f(x)=3x - 5, then f(2)=3·2 - 5=1. Treat f(x) like a machine that spits out answers when you feed it numbers. Online Math Learning: Notation & Evaluation
  9. Tackle real-world problems - Functions power up cost, revenue, and growth models. By combining cost functions, you can forecast total expenses and budget like a wizard. Applying these skills to real scenarios cements your understanding and saves the day in homework battles. Pearson: Real‑World Applications
  10. Quiz yourself with practice tests - Interactive problems and timed quizzes turbocharge your learning. Testing under pressure reveals gaps and builds confidence before the big exam. Dive into practice tests to see how far you've come and where you need a quick review. GreeneMath: Practice Test
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