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    Easy GRE Functions Questions Practice Questions

    July 8, 20268 min read63 views
    Easy GRE Functions Questions Practice Questions

    Concept Explanation

    Functions are mathematical rules that assign exactly one output value to each input value, often represented by the notation f(x)f(x). On the GRE, a function acts like a machine: you put a number (the input) into the machine, and it performs a specific set of operations defined by the equation to produce a result (the output). While functions can appear intimidating, Easy GRE Functions Questions typically focus on substitution, where you replace the variable in the function with a given number or expression. Understanding the relationship between the domain (inputs) and range (outputs) is fundamental for success in the GRE Prep process. Beyond simple substitution, you may encounter nested functions, such as f(g(x))f(g(x)), or defined operations where the test uses non-standard symbols like β—Š\Diamond or βŠ—\otimes to represent a specific sequence of arithmetic steps. Most introductory problems follow the linear model of f(x)=mx+bf(x) = mx + b, which is also a cornerstone of algebraic function theory. To improve your speed on these tasks, using a AI Question Generator can help you drill the mechanics of substitution until they become second nature.

    Solved Examples

    Review these three examples to understand how to approach functional notation and substitution on test day.

    1. Substitution Example: If f(x)=3x2βˆ’5x+2f(x) = 3x^2 - 5x + 2, what is the value of f(4)f(4)?
      1. Identify the input value, which is 4.
      2. Substitute 4 into the equation: f(4)=3(4)2βˆ’5(4)+2f(4) = 3(4)^2 - 5(4) + 2.
      3. Calculate the exponent first: 3(16)βˆ’5(4)+23(16) - 5(4) + 2.
      4. Perform multiplication: 48βˆ’20+248 - 20 + 2.
      5. Final calculation: 28+2=3028 + 2 = 30.
    2. Symbol-Defined Function: Let the operation aβŠ•ba \oplus b be defined by aβŠ•b=a+baba \oplus b = \frac{a + b}{ab}. Find the value of 2βŠ•32 \oplus 3.
      1. Treat aa as 2 and bb as 3.
      2. Plug the values into the given formula: 2+32Γ—3\frac{2 + 3}{2 \times 3}.
      3. Simplify the numerator: 2+3=52 + 3 = 5.
      4. Simplify the denominator: 2Γ—3=62 \times 3 = 6.
      5. The final result is 56\frac{5}{6}.
    3. Nested Functions: If f(x)=2x+1f(x) = 2x + 1 and g(x)=xβˆ’3g(x) = x - 3, find f(g(5))f(g(5)).
      1. Solve the inner function first: g(5)=5βˆ’3=2g(5) = 5 - 3 = 2.
      2. Now use the result (2) as the input for the outer function: f(2)f(2).
      3. Substitute 2 into f(x)f(x): 2(2)+12(2) + 1.
      4. Calculate: 4+1=54 + 1 = 5.

    Practice Questions

    Try these Easy GRE Functions Questions to build your confidence. You may find it helpful to compare these with Free GRE Practice Questions to see how function problems fit into the broader quantitative section.

    1. If f(x)=x3βˆ’10f(x) = x^3 - 10, what is the value of f(3)f(3)?
    2. For the function g(t)=5t+12g(t) = 5t + 12, if g(t)=37g(t) = 37, what is the value of tt?
    3. If h(x)=x+9h(x) = \sqrt{x + 9}, find h(16)h(16).

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    Practice GRE Questions
    1. Let f(x)=2xβˆ’7f(x) = 2x - 7. Find f(f(5))f(f(5)).
    2. If xβ‹„y=x2βˆ’yx \diamond y = x^2 - y, find the value of 4β‹„94 \diamond 9.
    3. If f(x)=12xf(x) = \frac{12}{x} and g(x)=x+4g(x) = x + 4, find f(g(2))f(g(2)).
    4. Given f(x)=4x+kf(x) = 4x + k, if f(2)=11f(2) = 11, what is the value of the constant kk?
    5. Which of the following values of xx makes f(x)=x2βˆ’4f(x) = x^2 - 4 equal to 0?
      • (A) 0
      • (B) 2
      • (C) 4
    6. If k(x)=7k(x) = 7 for all values of xx, find k(100)k(100).
    7. If f(x)=∣xβˆ’5∣f(x) = |x - 5|, find f(2)f(2).

    Answers & Explanations

    1. Answer: 17. Substitute 3 for xx: 33βˆ’10=27βˆ’10=173^3 - 10 = 27 - 10 = 17.
    2. Answer: 5. Set the equation equal to 37: 5t+12=375t + 12 = 37. Subtract 12 from both sides: 5t=255t = 25. Divide by 5: t=5t = 5.
    3. Answer: 5. Substitute 16 for xx: 16+9=25=5\sqrt{16 + 9} = \sqrt{25} = 5.
    4. Answer: -1. First find f(5)=2(5)βˆ’7=3f(5) = 2(5) - 7 = 3. Then find f(3)=2(3)βˆ’7=6βˆ’7=βˆ’1f(3) = 2(3) - 7 = 6 - 7 = -1.
    5. Answer: 7. Use the rule x2βˆ’yx^2 - y: 42βˆ’9=16βˆ’9=74^2 - 9 = 16 - 9 = 7.
    6. Answer: 2. Solve the inner function g(2)=2+4=6g(2) = 2 + 4 = 6. Then solve f(6)=126=2f(6) = \frac{12}{6} = 2.
    7. Answer: 3. Substitute x=2x=2 and f(2)=11f(2)=11: 4(2)+k=114(2) + k = 11 leads to 8+k=118 + k = 11, so k=3k = 3.
    8. Answer: 2. Setting x2βˆ’4=0x^2 - 4 = 0 gives x2=4x^2 = 4, so x=2x = 2 or βˆ’2-2. Option (B) is correct.
    9. Answer: 7. Since the function is a constant function, the output is always 7 regardless of the input.
    10. Answer: 3. Substitute 2 for xx: ∣2βˆ’5∣=βˆ£βˆ’3∣|2 - 5| = |-3|. The absolute value of -3 is 3.
    Interactive quizQuestion 1 of 5

    1. If \( f(x) = 5x - 3 \), what is \( f(2) \)?

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    Frequently Asked Questions

    What does the notation f(x) actually mean on the GRE?

    The notation f(x)f(x) represents the output of a function named ff when the input is xx. It is not a multiplication of ff and xx, but rather a way to describe the relationship between a dependent and independent variable.

    How do I handle strange symbols like circles or diamonds in function questions?

    These are called "defined operations" and they simply provide a custom rule for you to follow. Treat the symbol as a set of instructions, replacing the variables in the instruction with the numbers provided in the question.

    Can a function have more than one output for a single input?

    No, by definition, a mathematical function must assign exactly one output to each unique input. If you see a graph, you can use the vertical line test to confirm if it represents a function.

    What is the difference between domain and range?

    The domain refers to the set of all possible input values (usually xx) that the function can accept. The range refers to the set of all possible output values (usually yy or f(x)f(x)) that the function produces.

    Are nested functions common on the GRE?

    Yes, nested functions, also known as composite functions, appear frequently and require you to solve the innermost function first. You take the result of that inner calculation and use it as the new input for the outer function.

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