Study Function Definitions And Notation in ACT Math with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
All flashcards Flashcard 1: Identify the function that is constant for all inputs. Answer: f ( x ) = c f(x) = c f ( x ) = c , where c c c is a constant. A constant function outputs the same value regardless of input.
Flashcard 2: For f ( x ) = x x + 1 f(x) = \frac{x}{x+1} f ( x ) = x + 1 x , what is the vertical asymptote? Answer: Vertical asymptote is x = − 1 x = -1 x = − 1 . Vertical asymptotes occur where the denominator equals zero.
Flashcard 3: Simplify f ( x + 1 ) f(x+1) f ( x + 1 ) if f ( x ) = x 2 f(x)=x^2 f ( x ) = x 2 . Answer: f ( x + 1 ) = ( x + 1 ) 2 f(x+1)=(x+1)^2 f ( x + 1 ) = ( x + 1 ) 2 . Replace x x x with ( x + 1 ) (x+1) ( x + 1 ) in the function: f ( x + 1 ) = ( x + 1 ) 2 f(x+1)=(x+1)^2 f ( x + 1 ) = ( x + 1 ) 2 .
Flashcard 4: What is the x-intercept condition for y = f ( x ) y=f(x) y = f ( x ) in function notation? Answer: An x-intercept occurs where f ( x ) = 0 f(x)=0 f ( x ) = 0 . Solve for values where the function equals zero to find x-axis intersections.
Flashcard 5: What does the notation x ∈ domain ( f ) x\in\text{domain}(f) x ∈ domain ( f ) mean? Answer: x x x is an allowable input for f f f . The symbol ∈ \in ∈ means 'is an element of' or 'belongs to' the domain set.
Flashcard 6: What is f ( 0 ) f(0) f ( 0 ) for f ( x ) = 3 x 3 − 2 x f(x) = 3x^3 - 2x f ( x ) = 3 x 3 − 2 x ? Answer: f ( 0 ) = 0 f(0) = 0 f ( 0 ) = 0 . Substitute x = 0 x = 0 x = 0 : 3 ( 0 ) 3 − 2 ( 0 ) = 0 3(0)^3 - 2(0) = 0 3 ( 0 ) 3 − 2 ( 0 ) = 0 .
Flashcard 7: What does the notation f − 1 ( x ) f^{-1}(x) f − 1 ( x ) represent when it exists? Answer: The inverse function of f f f . The inverse function reverses the input-output relationship of the original function.
Flashcard 8: What is f ( 0 ) f(0) f ( 0 ) for f ( x ) = 5 x − 7 f(x) = 5x - 7 f ( x ) = 5 x − 7 ? Answer: f ( 0 ) = − 7 f(0) = -7 f ( 0 ) = − 7 . Substitute x = 0 x = 0 x = 0 : 5 ( 0 ) − 7 = − 7 5(0) - 7 = -7 5 ( 0 ) − 7 = − 7 .
Flashcard 9: For f ( x ) = x 3 − x f(x) = x^3 - x f ( x ) = x 3 − x , find f ( − 2 ) f(-2) f ( − 2 ) . Answer: f ( − 2 ) = − 6 f(-2) = -6 f ( − 2 ) = − 6 . Substitute x = − 2 x = -2 x = − 2 : ( − 2 ) 3 − ( − 2 ) = − 8 + 2 = − 6 (-2)^3 - (-2) = -8 + 2 = -6 ( − 2 ) 3 − ( − 2 ) = − 8 + 2 = − 6 .
Flashcard 10: What does the notation f ( x ) − f ( a ) f(x)-f(a) f ( x ) − f ( a ) represent? Answer: The difference between outputs at inputs x x x and a a a . This calculates the change in function values between two different input points.
Flashcard 11: If f ( x ) = 2 x + 1 f(x)=2x+1 f ( x ) = 2 x + 1 and g ( x ) = x 2 g(x)=x^2 g ( x ) = x 2 , find ( f ∘ g ) ( x ) (f\circ g)(x) ( f ∘ g ) ( x ) . Answer: ( f ∘ g ) ( x ) = 2 x 2 + 1 (f\circ g)(x)=2x^2+1 ( f ∘ g ) ( x ) = 2 x 2 + 1 . Substitute g ( x ) = x 2 g(x)=x^2 g ( x ) = x 2 into f f f : f ( g ( x ) ) = f ( x 2 ) = 2 ( x 2 ) + 1 = 2 x 2 + 1 f(g(x))=f(x^2)=2(x^2)+1=2x^2+1 f ( g ( x )) = f ( x 2 ) = 2 ( x 2 ) + 1 = 2 x 2 + 1 .
Flashcard 12: For f ( x ) = 1 x f(x) = \frac{1}{x} f ( x ) = x 1 , what is f ( 3 ) f(3) f ( 3 ) ? Answer: f ( 3 ) = 1 3 f(3) = \frac{1}{3} f ( 3 ) = 3 1 . Substitute x = 3 x = 3 x = 3 into the reciprocal function.
Flashcard 13: Define the zero of a function f ( x ) f(x) f ( x ) . Answer: Value of x x x where f ( x ) = 0 f(x) = 0 f ( x ) = 0 . A zero is an x-value that makes the function equal to zero.
Flashcard 14: Define the range of the function f ( x ) = x 2 f(x) = x^2 f ( x ) = x 2 . Answer: Non-negative real numbers. Squaring any real number produces a non-negative result.
Flashcard 15: If f ( x ) = 3 − x f(x)=3-x f ( x ) = 3 − x and g ( x ) = x 2 g(x)=x^2 g ( x ) = x 2 , find ( g ∘ f ) ( x ) (g\circ f)(x) ( g ∘ f ) ( x ) . Answer: ( g ∘ f ) ( x ) = ( 3 − x ) 2 (g\circ f)(x)=(3-x)^2 ( g ∘ f ) ( x ) = ( 3 − x ) 2 . Substitute f ( x ) = 3 − x f(x)=3-x f ( x ) = 3 − x into g g g : g ( f ( x ) ) = g ( 3 − x ) = ( 3 − x ) 2 g(f(x))=g(3-x)=(3-x)^2 g ( f ( x )) = g ( 3 − x ) = ( 3 − x ) 2 .
Flashcard 16: If f ( x ) = x + 1 2 f(x)=\frac{x+1}{2} f ( x ) = 2 x + 1 , what is f − 1 ( x ) f^{-1}(x) f − 1 ( x ) ? Answer: f − 1 ( x ) = 2 x − 1 f^{-1}(x)=2x-1 f − 1 ( x ) = 2 x − 1 . Swap x x x and y y y , then solve for y y y : x = y + 1 2 x=\frac{y+1}{2} x = 2 y + 1 gives y = 2 x − 1 y=2x-1 y = 2 x − 1 .
Flashcard 17: Find f ( a ) f(a) f ( a ) if f ( x ) = 3 x 2 − 2 x f(x)=3x^2-2x f ( x ) = 3 x 2 − 2 x . Answer: f ( a ) = 3 a 2 − 2 a f(a)=3a^2-2a f ( a ) = 3 a 2 − 2 a . Replace every occurrence of x x x with a a a in the function expression.
Flashcard 18: For f ( x ) = 1 x − 1 f(x) = \frac{1}{x-1} f ( x ) = x − 1 1 , what is the domain? Answer: All real numbers except x = 1 x = 1 x = 1 . Division by zero occurs when x = 1 x = 1 x = 1 .
Flashcard 19: Define the range of the function f ( x ) = x 2 f(x) = x^2 f ( x ) = x 2 . Answer: Non-negative real numbers. Squaring any real number produces a non-negative result.
Flashcard 20: Identify the range of f ( x ) = 1 x f(x) = \frac{1}{x} f ( x ) = x 1 . Answer: All real numbers except 0 0 0 . The reciprocal function cannot equal zero for any real input.
Flashcard 21: What does the notation f ( x ) f(x) f ( x ) represent in a function? Answer: f ( x ) f(x) f ( x ) is the output value of f f f when the input is x x x . This defines function notation where f f f is the function name and x x x is the input variable.
Flashcard 22: What does the notation f ( x ) + g ( x ) f(x)+g(x) f ( x ) + g ( x ) mean? Answer: A new function whose output is f ( x ) + g ( x ) f(x)+g(x) f ( x ) + g ( x ) . Function addition creates a new function by adding corresponding outputs pointwise.
Flashcard 23: Evaluate f ( − 2 ) f(-2) f ( − 2 ) if f ( x ) = x 2 + 4 f(x)=x^2+4 f ( x ) = x 2 + 4 . Answer: f ( − 2 ) = 8 f(-2)=8 f ( − 2 ) = 8 . Substitute x = − 2 x=-2 x = − 2 : f ( − 2 ) = ( − 2 ) 2 + 4 = 4 + 4 = 8 f(-2)=(-2)^2+4=4+4=8 f ( − 2 ) = ( − 2 ) 2 + 4 = 4 + 4 = 8 .
Flashcard 24: What is the range of a function, in words? Answer: The set of all possible output values. Range consists of all y y y -values that the function can actually produce as outputs.
Flashcard 25: What is the correct interpretation of f − 1 ( x ) f^{-1}(x) f − 1 ( x ) (not a power)? Answer: f − 1 ( x ) f^{-1}(x) f − 1 ( x ) is the inverse function, not 1 f ( x ) \frac{1}{f(x)} f ( x ) 1 . The superscript − 1 -1 − 1 denotes inverse function, not reciprocal or negative exponent.
Flashcard 26: What is the domain of a function, in words? Answer: The set of all allowable input values. Domain consists of all x x x -values where the function is defined and produces real outputs.
Flashcard 27: What is the domain of f ( x ) = 1 x 2 − 4 f(x) = \frac{1}{x^2 - 4} f ( x ) = x 2 − 4 1 ? Answer: All real numbers except x = ± 2 x = \text{±}2 x = ± 2 . Denominator equals zero when x 2 = 4 x^2 = 4 x 2 = 4 , so x = ± 2 x = ±2 x = ± 2 .
Flashcard 28: What is f ( − 3 ) f(-3) f ( − 3 ) for f ( x ) = 2 x + 1 f(x) = 2x + 1 f ( x ) = 2 x + 1 ? Answer: f ( − 3 ) = − 5 f(-3) = -5 f ( − 3 ) = − 5 . Substitute x = − 3 x = -3 x = − 3 : 2 ( − 3 ) + 1 = − 6 + 1 = − 5 2(-3) + 1 = -6 + 1 = -5 2 ( − 3 ) + 1 = − 6 + 1 = − 5 .
Flashcard 29: What does the notation f ( x ) − f ( a ) f(x)-f(a) f ( x ) − f ( a ) represent? Answer: The difference between outputs at inputs x x x and a a a . This calculates the change in function values between two different input points.
Flashcard 30: What is the defining relationship between a function and its inverse? Answer: f ( f − 1 ( x ) ) = x f(f^{-1}(x))=x f ( f − 1 ( x )) = x and f − 1 ( f ( x ) ) = x f^{-1}(f(x))=x f − 1 ( f ( x )) = x . These compositions demonstrate that inverse functions undo each other's operations.
Flashcard 31: What is the meaning of the equation f ( x ) = 0 f(x)=0 f ( x ) = 0 ? Answer: The input values x x x for which the output equals 0 0 0 . These are the zeros or roots of the function, where the graph crosses the x-axis.
Flashcard 32: What does the notation f ( 2 ) = 7 f(2) = 7 f ( 2 ) = 7 indicate? Answer: The output is 7 7 7 when input is 2 2 2 . Function notation shows the input-output relationship.
Flashcard 33: What is the domain of a function, in words? Answer: The set of all allowable input values. Domain consists of all x x x -values where the function is defined and produces real outputs.
Flashcard 34: What is the meaning of the equation f ( x ) = 0 f(x)=0 f ( x ) = 0 ? Answer: The input values x x x for which the output equals 0 0 0 . These are the zeros or roots of the function, where the graph crosses the x-axis.
Flashcard 35: If f ( x ) = x 2 f(x)=x^2 f ( x ) = x 2 and g ( x ) = x − 1 g(x)=x-1 g ( x ) = x − 1 , find ( g ∘ f ) ( 2 ) (g\circ f)(2) ( g ∘ f ) ( 2 ) . Answer: ( g ∘ f ) ( 2 ) = 3 (g\circ f)(2)=3 ( g ∘ f ) ( 2 ) = 3 . First find f ( 2 ) = 4 f(2)=4 f ( 2 ) = 4 , then g ( 4 ) = 4 − 1 = 3 g(4)=4-1=3 g ( 4 ) = 4 − 1 = 3 .
Flashcard 36: For f ( x ) = x 2 f(x) = x^2 f ( x ) = x 2 , what is the vertex of the parabola? Answer: Vertex is ( 0 , 0 ) (0, 0) ( 0 , 0 ) . The vertex of x 2 x^2 x 2 occurs at the origin where the derivative is zero.
Flashcard 37: What is the y-intercept of a function y = f ( x ) y=f(x) y = f ( x ) in function notation? Answer: The y-intercept is f ( 0 ) f(0) f ( 0 ) . Set x = 0 x=0 x = 0 and evaluate the function to find where the graph crosses the y-axis.
Flashcard 38: What is the standard meaning of the statement y = f ( x ) y=f(x) y = f ( x ) ? Answer: y y y equals the output of function f f f for input x x x . This is the standard function equation format relating dependent and independent variables.
Flashcard 39: What does the notation y ∈ range ( f ) y\in\text{range}(f) y ∈ range ( f ) mean? Answer: y y y is a possible output value of f f f . The symbol ∈ \in ∈ means the output value exists within the function's range set.
Flashcard 40: If f ( x ) = x 2 f(x) = x^2 f ( x ) = x 2 and g ( x ) = 2 x g(x) = 2x g ( x ) = 2 x , find ( f ∘ g ) ( x ) (f \, \text{∘} \, g)(x) ( f ∘ g ) ( x ) . Answer: ( f ∘ g ) ( x ) = 4 x 2 (f \, \text{∘} \, g)(x) = 4x^2 ( f ∘ g ) ( x ) = 4 x 2 . First apply g g g , then f f f : f ( g ( x ) ) = f ( 2 x ) = ( 2 x ) 2 = 4 x 2 f(g(x)) = f(2x) = (2x)^2 = 4x^2 f ( g ( x )) = f ( 2 x ) = ( 2 x ) 2 = 4 x 2 .
Flashcard 41: What does the notation f ( 2 ) = 7 f(2) = 7 f ( 2 ) = 7 indicate? Answer: The output is 7 7 7 when input is 2 2 2 . Function notation shows the input-output relationship.
Flashcard 42: What is the domain restriction for f ( x ) = x + 5 f(x)=\sqrt{x+5} f ( x ) = x + 5 (real numbers)? Answer: Domain: x ≥ − 5 x\ge -5 x ≥ − 5 . The radicand x + 5 x+5 x + 5 must be non-negative for real square root values.
Flashcard 43: What does the notation x ∈ domain ( f ) x\in\text{domain}(f) x ∈ domain ( f ) mean? Answer: x x x is an allowable input for f f f . The symbol ∈ \in ∈ means 'is an element of' or 'belongs to' the domain set.
Flashcard 44: What does the notation f ( x + h ) f(x+h) f ( x + h ) indicate compared with f ( x ) f(x) f ( x ) ? Answer: The output when the input is shifted to x + h x+h x + h . The input is horizontally shifted by h h h units, affecting the function's evaluation point.
Flashcard 45: What is the defining relationship between a function and its inverse? Answer: f ( f − 1 ( x ) ) = x f(f^{-1}(x))=x f ( f − 1 ( x )) = x and f − 1 ( f ( x ) ) = x f^{-1}(f(x))=x f − 1 ( f ( x )) = x . These compositions demonstrate that inverse functions undo each other's operations.
Flashcard 46: Find x x x if f ( x ) = x 2 − 9 f(x)=x^2-9 f ( x ) = x 2 − 9 and f ( x ) = 0 f(x)=0 f ( x ) = 0 . Answer: x = 3 x=3 x = 3 or x = − 3 x=-3 x = − 3 . Set x 2 − 9 = 0 x^2-9=0 x 2 − 9 = 0 , so x 2 = 9 x^2=9 x 2 = 9 , giving x = ± 3 x=\pm 3 x = ± 3 .
Flashcard 47: What does it mean if f ( a ) ≠ f ( b ) f(a)\neq f(b) f ( a ) = f ( b ) ? Answer: Inputs a a a and b b b produce different outputs. This indicates the function produces distinct outputs for these different inputs.
Flashcard 48: What is f ( 1 ) f(1) f ( 1 ) for f ( x ) = 3 x + 5 f(x) = 3x + 5 f ( x ) = 3 x + 5 ? Answer: f ( 1 ) = 8 f(1) = 8 f ( 1 ) = 8 . Substitute x = 1 x = 1 x = 1 : 3 ( 1 ) + 5 = 8 3(1) + 5 = 8 3 ( 1 ) + 5 = 8 .
Flashcard 49: If f ( x ) = x + 1 2 f(x)=\frac{x+1}{2} f ( x ) = 2 x + 1 , what is f − 1 ( x ) f^{-1}(x) f − 1 ( x ) ? Answer: f − 1 ( x ) = 2 x − 1 f^{-1}(x)=2x-1 f − 1 ( x ) = 2 x − 1 . Swap x x x and y y y , then solve for y y y : x = y + 1 2 x=\frac{y+1}{2} x = 2 y + 1 gives y = 2 x − 1 y=2x-1 y = 2 x − 1 .
Flashcard 50: Evaluate f ( 0 ) f(0) f ( 0 ) if f ( x ) = x − 1 x + 2 f(x)=\frac{x-1}{x+2} f ( x ) = x + 2 x − 1 . Answer: f ( 0 ) = − 1 2 f(0)=-\frac{1}{2} f ( 0 ) = − 2 1 . Substitute x = 0 x=0 x = 0 : f ( 0 ) = 0 − 1 0 + 2 = − 1 2 = − 1 2 f(0)=\frac{0-1}{0+2}=\frac{-1}{2}=-\frac{1}{2} f ( 0 ) = 0 + 2 0 − 1 = 2 − 1 = − 2 1 .
Flashcard 51: If f ( x ) = 3 − x f(x)=3-x f ( x ) = 3 − x and g ( x ) = x 2 g(x)=x^2 g ( x ) = x 2 , find ( g ∘ f ) ( x ) (g\circ f)(x) ( g ∘ f ) ( x ) . Answer: ( g ∘ f ) ( x ) = ( 3 − x ) 2 (g\circ f)(x)=(3-x)^2 ( g ∘ f ) ( x ) = ( 3 − x ) 2 . Substitute f ( x ) = 3 − x f(x)=3-x f ( x ) = 3 − x into g g g : g ( f ( x ) ) = g ( 3 − x ) = ( 3 − x ) 2 g(f(x))=g(3-x)=(3-x)^2 g ( f ( x )) = g ( 3 − x ) = ( 3 − x ) 2 .
Flashcard 52: What is f ( 3 ) f(3) f ( 3 ) if f ( x ) = x 2 − 4 x + 4 f(x) = x^2 - 4x + 4 f ( x ) = x 2 − 4 x + 4 ? Answer: f ( 3 ) = 1 f(3) = 1 f ( 3 ) = 1 . Substitute x = 3 x = 3 x = 3 : ( 3 ) 2 − 4 ( 3 ) + 4 = 9 − 12 + 4 = 1 (3)^2 - 4(3) + 4 = 9 - 12 + 4 = 1 ( 3 ) 2 − 4 ( 3 ) + 4 = 9 − 12 + 4 = 1 .
Flashcard 53: What does the notation y ∈ range ( f ) y\in\text{range}(f) y ∈ range ( f ) mean? Answer: y y y is a possible output value of f f f . The symbol ∈ \in ∈ means the output value exists within the function's range set.
Flashcard 54: For f ( x ) = 3 x + 2 f(x) = 3x + 2 f ( x ) = 3 x + 2 , what is f ( 4 ) f(4) f ( 4 ) ? Answer: f ( 4 ) = 14 f(4) = 14 f ( 4 ) = 14 . Substitute x = 4 x = 4 x = 4 into the function: 3 ( 4 ) + 2 = 14 3(4) + 2 = 14 3 ( 4 ) + 2 = 14 .
Flashcard 55: What does it mean if f ( a ) ≠ f ( b ) f(a)\neq f(b) f ( a ) = f ( b ) ? Answer: Inputs a a a and b b b produce different outputs. This indicates the function produces distinct outputs for these different inputs.
Flashcard 56: Evaluate f ( 0 ) f(0) f ( 0 ) if f ( x ) = x − 1 x + 2 f(x)=\frac{x-1}{x+2} f ( x ) = x + 2 x − 1 . Answer: f ( 0 ) = − 1 2 f(0)=-\frac{1}{2} f ( 0 ) = − 2 1 . Substitute x = 0 x=0 x = 0 : f ( 0 ) = 0 − 1 0 + 2 = − 1 2 = − 1 2 f(0)=\frac{0-1}{0+2}=\frac{-1}{2}=-\frac{1}{2} f ( 0 ) = 0 + 2 0 − 1 = 2 − 1 = − 2 1 .
Flashcard 57: Simplify f ( x ) − f ( 1 ) f(x)-f(1) f ( x ) − f ( 1 ) if f ( x ) = x 2 + 2 x f(x)=x^2+2x f ( x ) = x 2 + 2 x . Answer: f ( x ) − f ( 1 ) = x 2 + 2 x − 3 f(x)-f(1)=x^2+2x-3 f ( x ) − f ( 1 ) = x 2 + 2 x − 3 . Calculate f ( x ) = ( x 2 + 2 x ) f(x)=(x^2+2x) f ( x ) = ( x 2 + 2 x ) and f ( 1 ) = 3 f(1)=3 f ( 1 ) = 3 , so the difference is x 2 + 2 x − 3 x^2+2x-3 x 2 + 2 x − 3 .
Flashcard 58: What does f ( a ) f(a) f ( a ) represent if f ( x ) f(x) f ( x ) is a function? Answer: Value of f ( x ) f(x) f ( x ) at x = a x = a x = a . Function notation indicates the output when input is a a a .
Flashcard 59: What is the slope of f ( x ) = 7 x − 4 f(x) = 7x - 4 f ( x ) = 7 x − 4 ? Answer: Slope is 7 7 7 . Linear functions have the form m x + b mx + b m x + b where m m m is the slope.
Flashcard 60: Evaluate f ( 3 ) f(3) f ( 3 ) if f ( x ) = 2 x − 5 f(x)=2x-5 f ( x ) = 2 x − 5 . Answer: f ( 3 ) = 1 f(3)=1 f ( 3 ) = 1 . Substitute x = 3 x=3 x = 3 : f ( 3 ) = 2 ( 3 ) − 5 = 6 − 5 = 1 f(3)=2(3)-5=6-5=1 f ( 3 ) = 2 ( 3 ) − 5 = 6 − 5 = 1 .
Flashcard 61: Identify the notation for the reciprocal of a function output. Answer: 1 f ( x ) \frac{1}{f(x)} f ( x ) 1 . This represents one divided by the function output, the multiplicative inverse.
Flashcard 62: Simplify f ( x ) − f ( 1 ) f(x)-f(1) f ( x ) − f ( 1 ) if f ( x ) = x 2 + 2 x f(x)=x^2+2x f ( x ) = x 2 + 2 x . Answer: f ( x ) − f ( 1 ) = x 2 + 2 x − 3 f(x)-f(1)=x^2+2x-3 f ( x ) − f ( 1 ) = x 2 + 2 x − 3 . Calculate f ( x ) = ( x 2 + 2 x ) f(x)=(x^2+2x) f ( x ) = ( x 2 + 2 x ) and f ( 1 ) = 3 f(1)=3 f ( 1 ) = 3 , so the difference is x 2 + 2 x − 3 x^2+2x-3 x 2 + 2 x − 3 .
Flashcard 63: What does the notation ∣ f ( x ) ∣ |f(x)| ∣ f ( x ) ∣ mean? Answer: The absolute value of the output f ( x ) f(x) f ( x ) . Takes the absolute value of whatever the function outputs for input x x x .
Flashcard 64: What does the notation ( f ∘ g ) ( x ) (f\circ g)(x) ( f ∘ g ) ( x ) mean? Answer: ( f ∘ g ) ( x ) = f ( g ( x ) ) (f\circ g)(x)=f(g(x)) ( f ∘ g ) ( x ) = f ( g ( x )) . Composition means applying g g g first, then applying f f f to that result.
Flashcard 65: Find the y-intercept of f ( x ) = x 2 + 3 x + 2 f(x) = x^2 + 3x + 2 f ( x ) = x 2 + 3 x + 2 . Answer: The point ( 0 , 2 ) (0, 2) ( 0 , 2 ) . Set x = 0 x = 0 x = 0 to find where the graph crosses the y-axis.
Flashcard 66: For f ( x ) = 1 x − 1 f(x) = \frac{1}{x-1} f ( x ) = x − 1 1 , what is the domain? Answer: All real numbers except x = 1 x = 1 x = 1 . Division by zero occurs when x = 1 x = 1 x = 1 .
Flashcard 67: What is f ( 0 ) f(0) f ( 0 ) for f ( x ) = 3 x 3 − 2 x f(x) = 3x^3 - 2x f ( x ) = 3 x 3 − 2 x ? Answer: f ( 0 ) = 0 f(0) = 0 f ( 0 ) = 0 . Substitute x = 0 x = 0 x = 0 : 3 ( 0 ) 3 − 2 ( 0 ) = 0 3(0)^3 - 2(0) = 0 3 ( 0 ) 3 − 2 ( 0 ) = 0 .
Flashcard 68: Find the inverse of f ( x ) = 2 x + 3 f(x) = 2x + 3 f ( x ) = 2 x + 3 . Answer: f − 1 ( x ) = x − 3 2 f^{-1}(x) = \frac{x-3}{2} f − 1 ( x ) = 2 x − 3 . Solve y = 2 x + 3 y = 2x + 3 y = 2 x + 3 for x x x , then swap variables.
Flashcard 69: What is the range of f ( x ) = abs ( x ) f(x) = \text{abs}(x) f ( x ) = abs ( x ) ? Answer: Non-negative real numbers. Absolute value function always produces non-negative outputs.
Flashcard 70: Find f ( 2 ) f(2) f ( 2 ) if f ( x ) = 2 x 2 − 3 x + 1 f(x) = 2x^2 - 3x + 1 f ( x ) = 2 x 2 − 3 x + 1 . Answer: f ( 2 ) = 3 f(2) = 3 f ( 2 ) = 3 . Substitute x = 2 x = 2 x = 2 : 2 ( 4 ) − 3 ( 2 ) + 1 = 8 − 6 + 1 = 3 2(4) - 3(2) + 1 = 8 - 6 + 1 = 3 2 ( 4 ) − 3 ( 2 ) + 1 = 8 − 6 + 1 = 3 .
Flashcard 71: What is the value of f ( − 5 ) f(-5) f ( − 5 ) if f ( x ) = − x + 6 f(x) = -x + 6 f ( x ) = − x + 6 ? Answer: f ( − 5 ) = 11 f(-5) = 11 f ( − 5 ) = 11 . Substitute x = − 5 x = -5 x = − 5 : − ( − 5 ) + 6 = 5 + 6 = 11 -(-5) + 6 = 5 + 6 = 11 − ( − 5 ) + 6 = 5 + 6 = 11 .
Flashcard 72: What is the notation for the composition of f f f and g g g ? Answer: ( f ∘ g ) ( x ) (f \, \text{∘} \, g)(x) ( f ∘ g ) ( x ) . This symbol represents function composition, where one function is applied to another.
Flashcard 73: What is the notation for the composition of f f f and g g g ? Answer: ( f ∘ g ) ( x ) (f \, \text{∘} \, g)(x) ( f ∘ g ) ( x ) . This symbol represents function composition, where one function is applied to another.
Flashcard 74: What does the notation f ( x ) = x f(x)=\sqrt{x} f ( x ) = x imply about the domain (real numbers)? Answer: Domain: x ≥ 0 x\ge 0 x ≥ 0 . Square root requires non-negative inputs to produce real number outputs.
Flashcard 75: Express f ( x ) = x 2 + 2 x + 1 f(x) = x^2 + 2x + 1 f ( x ) = x 2 + 2 x + 1 in vertex form. Answer: f ( x ) = ( x + 1 ) 2 f(x) = (x+1)^2 f ( x ) = ( x + 1 ) 2 . Factor the perfect square trinomial to vertex form.
Flashcard 76: What does the notation f ( x ) f(x) f ( x ) represent in a function? Answer: f ( x ) f(x) f ( x ) is the output value of f f f when the input is x x x . This defines function notation where f f f is the function name and x x x is the input variable.
Flashcard 77: Identify the range of f ( x ) = 1 x f(x) = \frac{1}{x} f ( x ) = x 1 . Answer: All real numbers except 0 0 0 . The reciprocal function cannot equal zero for any real input.
Flashcard 78: If f ( x ) = x − 5 f(x)=x-5 f ( x ) = x − 5 , what is f − 1 ( x ) f^{-1}(x) f − 1 ( x ) ? Answer: f − 1 ( x ) = x + 5 f^{-1}(x)=x+5 f − 1 ( x ) = x + 5 . For linear functions, the inverse reverses the operation: subtract becomes add.
Flashcard 79: What is f ( 3 ) f(3) f ( 3 ) if f ( x ) = x 2 − 4 x + 4 f(x) = x^2 - 4x + 4 f ( x ) = x 2 − 4 x + 4 ? Answer: f ( 3 ) = 1 f(3) = 1 f ( 3 ) = 1 . Substitute x = 3 x = 3 x = 3 : ( 3 ) 2 − 4 ( 3 ) + 4 = 9 − 12 + 4 = 1 (3)^2 - 4(3) + 4 = 9 - 12 + 4 = 1 ( 3 ) 2 − 4 ( 3 ) + 4 = 9 − 12 + 4 = 1 .
Flashcard 80: For f ( x ) = x x + 1 f(x) = \frac{x}{x+1} f ( x ) = x + 1 x , what is the vertical asymptote? Answer: Vertical asymptote is x = − 1 x = -1 x = − 1 . Vertical asymptotes occur where the denominator equals zero.
Flashcard 81: Simplify f ( 2 x ) f(2x) f ( 2 x ) if f ( x ) = x 2 − 3 f(x)=x^2-3 f ( x ) = x 2 − 3 . Answer: f ( 2 x ) = 4 x 2 − 3 f(2x)=4x^2-3 f ( 2 x ) = 4 x 2 − 3 . Replace x x x with 2 x 2x 2 x : f ( 2 x ) = ( 2 x ) 2 − 3 = 4 x 2 − 3 f(2x)=(2x)^2-3=4x^2-3 f ( 2 x ) = ( 2 x ) 2 − 3 = 4 x 2 − 3 .
Flashcard 82: Find f ( 2 ) f(2) f ( 2 ) if f ( x ) = 2 x 2 − 3 x + 1 f(x) = 2x^2 - 3x + 1 f ( x ) = 2 x 2 − 3 x + 1 . Answer: f ( 2 ) = 3 f(2) = 3 f ( 2 ) = 3 . Substitute x = 2 x = 2 x = 2 : 2 ( 4 ) − 3 ( 2 ) + 1 = 8 − 6 + 1 = 3 2(4) - 3(2) + 1 = 8 - 6 + 1 = 3 2 ( 4 ) − 3 ( 2 ) + 1 = 8 − 6 + 1 = 3 .
Flashcard 83: Simplify f ( x ) = x 2 − 1 x − 1 f(x) = \frac{x^2 - 1}{x - 1} f ( x ) = x − 1 x 2 − 1 . Answer: f ( x ) = x + 1 f(x) = x + 1 f ( x ) = x + 1 , x ≠ 1 x \neq 1 x = 1 . Factor and cancel, but note the domain restriction.
Flashcard 84: What is the slope of f ( x ) = 7 x − 4 f(x) = 7x - 4 f ( x ) = 7 x − 4 ? Answer: Slope is 7 7 7 . Linear functions have the form m x + b mx + b m x + b where m m m is the slope.
Flashcard 85: If f ( x ) = x + 4 f(x)=x+4 f ( x ) = x + 4 and g ( x ) = 2 x g(x)=2x g ( x ) = 2 x , find ( f ∘ g ) ( 3 ) (f\circ g)(3) ( f ∘ g ) ( 3 ) . Answer: ( f ∘ g ) ( 3 ) = 10 (f\circ g)(3)=10 ( f ∘ g ) ( 3 ) = 10 . First find g ( 3 ) = 6 g(3)=6 g ( 3 ) = 6 , then f ( 6 ) = 6 + 4 = 10 f(6)=6+4=10 f ( 6 ) = 6 + 4 = 10 .
Flashcard 86: Find f ( x + 1 ) f(x+1) f ( x + 1 ) for f ( x ) = x 2 − x f(x) = x^2 - x f ( x ) = x 2 − x . Answer: f ( x + 1 ) = x 2 + x f(x+1) = x^2 + x f ( x + 1 ) = x 2 + x . Replace x x x with x + 1 x+1 x + 1 in the function: ( x + 1 ) 2 − ( x + 1 ) (x+1)^2 - (x+1) ( x + 1 ) 2 − ( x + 1 ) .
Flashcard 87: What is the notation for the absolute value of x x x ? Answer: ∣ x ∣ |x| ∣ x ∣ . Vertical bars around a variable indicate absolute value.
Flashcard 88: What does the notation f − 1 ( x ) f^{-1}(x) f − 1 ( x ) represent when it exists? Answer: The inverse function of f f f . The inverse function reverses the input-output relationship of the original function.
Flashcard 89: What does it mean if f ( a ) = f ( b ) f(a)=f(b) f ( a ) = f ( b ) ? Answer: Inputs a a a and b b b produce the same output. Different inputs can map to the same output value in many-to-one relationships.
Flashcard 90: What is f ( − 3 ) f(-3) f ( − 3 ) for f ( x ) = 2 x + 1 f(x) = 2x + 1 f ( x ) = 2 x + 1 ? Answer: f ( − 3 ) = − 5 f(-3) = -5 f ( − 3 ) = − 5 . Substitute x = − 3 x = -3 x = − 3 : 2 ( − 3 ) + 1 = − 6 + 1 = − 5 2(-3) + 1 = -6 + 1 = -5 2 ( − 3 ) + 1 = − 6 + 1 = − 5 .
Flashcard 91: What is the domain restriction for f ( x ) = x + 5 f(x)=\sqrt{x+5} f ( x ) = x + 5 (real numbers)? Answer: Domain: x ≥ − 5 x\ge -5 x ≥ − 5 . The radicand x + 5 x+5 x + 5 must be non-negative for real square root values.
Flashcard 92: What is the domain of f ( x ) = 1 x 2 − 4 f(x) = \frac{1}{x^2 - 4} f ( x ) = x 2 − 4 1 ? Answer: All real numbers except x = ± 2 x = \text{±}2 x = ± 2 . Denominator equals zero when x 2 = 4 x^2 = 4 x 2 = 4 , so x = ± 2 x = ±2 x = ± 2 .
Flashcard 93: Find x x x if f ( x ) = x 2 − 9 f(x)=x^2-9 f ( x ) = x 2 − 9 and f ( x ) = 0 f(x)=0 f ( x ) = 0 . Answer: x = 3 x=3 x = 3 or x = − 3 x=-3 x = − 3 . Set x 2 − 9 = 0 x^2-9=0 x 2 − 9 = 0 , so x 2 = 9 x^2=9 x 2 = 9 , giving x = ± 3 x=\pm 3 x = ± 3 .
Flashcard 94: What is the domain restriction for f ( x ) = 1 x − 3 f(x)=\frac{1}{x-3} f ( x ) = x − 3 1 (real numbers)? Answer: Domain: x ≠ 3 x\ne 3 x = 3 . Set the denominator x − 3 ≠ 0 x-3\neq 0 x − 3 = 0 to avoid division by zero.
Flashcard 95: What is the domain restriction for f ( x ) = 1 x − 3 f(x)=\frac{1}{x-3} f ( x ) = x − 3 1 (real numbers)? Answer: Domain: x ≠ 3 x\ne 3 x = 3 . Set the denominator x − 3 ≠ 0 x-3\neq 0 x − 3 = 0 to avoid division by zero.
Flashcard 96: What is the correct interpretation of f − 1 ( x ) f^{-1}(x) f − 1 ( x ) (not a power)? Answer: f − 1 ( x ) f^{-1}(x) f − 1 ( x ) is the inverse function, not 1 f ( x ) \frac{1}{f(x)} f ( x ) 1 . The superscript − 1 -1 − 1 denotes inverse function, not reciprocal or negative exponent.
Flashcard 97: Identify the notation for the reciprocal of a function output. Answer: 1 f ( x ) \frac{1}{f(x)} f ( x ) 1 . This represents one divided by the function output, the multiplicative inverse.
Flashcard 98: Simplify f ( x + 1 ) f(x+1) f ( x + 1 ) if f ( x ) = x 2 f(x)=x^2 f ( x ) = x 2 . Answer: f ( x + 1 ) = ( x + 1 ) 2 f(x+1)=(x+1)^2 f ( x + 1 ) = ( x + 1 ) 2 . Replace x x x with ( x + 1 ) (x+1) ( x + 1 ) in the function: f ( x + 1 ) = ( x + 1 ) 2 f(x+1)=(x+1)^2 f ( x + 1 ) = ( x + 1 ) 2 .
Flashcard 99: Find f ( x + 1 ) f(x+1) f ( x + 1 ) for f ( x ) = x 2 − x f(x) = x^2 - x f ( x ) = x 2 − x . Answer: f ( x + 1 ) = x 2 + x f(x+1) = x^2 + x f ( x + 1 ) = x 2 + x . Replace x x x with x + 1 x+1 x + 1 in the function: ( x + 1 ) 2 − ( x + 1 ) (x+1)^2 - (x+1) ( x + 1 ) 2 − ( x + 1 ) .
Flashcard 100: Evaluate f ( − 2 ) f(-2) f ( − 2 ) if f ( x ) = x 2 + 4 f(x)=x^2+4 f ( x ) = x 2 + 4 . Answer: f ( − 2 ) = 8 f(-2)=8 f ( − 2 ) = 8 . Substitute x = − 2 x=-2 x = − 2 : f ( − 2 ) = ( − 2 ) 2 + 4 = 4 + 4 = 8 f(-2)=(-2)^2+4=4+4=8 f ( − 2 ) = ( − 2 ) 2 + 4 = 4 + 4 = 8 .