Study Function Notation And Evaluation in Algebra with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
All flashcards Flashcard 1: What does it mean if an input x x x is not in the domain of f f f ? Answer: f ( x ) f(x) f ( x ) is undefined for that input. The function cannot be evaluated at that input.
Flashcard 2: Identify whether f ( 2 ) f(2) f ( 2 ) is defined if f ( x ) = 1 x − 2 f(x)=\frac{1}{x-2} f ( x ) = x − 2 1 . Answer: f ( 2 ) f(2) f ( 2 ) is undefined. Division by zero occurs when x = 2 x=2 x = 2 .
Flashcard 3: What is the domain of f ( x ) = 1 x + 3 f(x)=\frac{1}{x+3} f ( x ) = x + 3 1 in set notation? Answer: All real x x x such that x ≠ − 3 x\neq -3 x = − 3 . Denominator cannot equal zero, so exclude x = − 3 x=-3 x = − 3 .
Flashcard 4: What is the domain of a function in words? Answer: The set of all allowable input values. Only these values can be substituted into the function.
Flashcard 5: What is the domain of f ( x ) = x − 5 f(x)=\sqrt{x-5} f ( x ) = x − 5 in inequality form? Answer: x ≥ 5 x\ge 5 x ≥ 5 . Square root requires non-negative radicand.
Flashcard 6: Find the domain of f ( x ) = 2 x f(x)=\frac{2}{x} f ( x ) = x 2 in restriction form. Answer: All real x x x such that x ≠ 0 x\neq 0 x = 0 . Denominator cannot equal zero.
Flashcard 7: Interpret P ( 5 ) = 1200 P(5)=1200 P ( 5 ) = 1200 if P ( t ) P(t) P ( t ) is population after t t t years. Answer: After 5 5 5 years, the population is 1200 1200 1200 . Function output shows population at given time.
Flashcard 8: Find p ( x − 1 ) p(x-1) p ( x − 1 ) if p ( x ) = x 2 p(x)=x^2 p ( x ) = x 2 . Answer: p ( x − 1 ) = ( x − 1 ) 2 p(x-1)=(x-1)^2 p ( x − 1 ) = ( x − 1 ) 2 . Replace x x x with ( x − 1 ) (x-1) ( x − 1 ) : p ( x − 1 ) = ( x − 1 ) 2 p(x-1)=(x-1)^2 p ( x − 1 ) = ( x − 1 ) 2 .
Flashcard 9: Find f ( − 2 ) f(-2) f ( − 2 ) if f ( x ) = − x + 6 f(x)=-x+6 f ( x ) = − x + 6 . Answer: f ( − 2 ) = 8 f(-2)=8 f ( − 2 ) = 8 . Substitute x = − 2 x=-2 x = − 2 : f ( − 2 ) = − ( − 2 ) + 6 = 8 f(-2)=-(-2)+6=8 f ( − 2 ) = − ( − 2 ) + 6 = 8 .
Flashcard 10: What does the statement f ( x ) = 0 f(x)=0 f ( x ) = 0 ask you to find? Answer: All inputs x x x that make the output equal to 0 0 0 . Find the zeros or roots of the function.
Flashcard 11: What is the difference between f ( x ) f(x) f ( x ) and f x f\,x f x in algebra? Answer: f ( x ) f(x) f ( x ) is a function value; f x f\,x f x is multiplication. Parentheses indicate function evaluation, not multiplication.
Flashcard 12: Find q ( − 2 ) q(-2) q ( − 2 ) if q ( x ) = x 2 + 4 x + 1 q(x)=x^2+4x+1 q ( x ) = x 2 + 4 x + 1 . Answer: q ( − 2 ) = − 3 q(-2)=-3 q ( − 2 ) = − 3 . Substitute x = − 2 x=-2 x = − 2 : q ( − 2 ) = ( − 2 ) 2 + 4 ( − 2 ) + 1 = − 3 q(-2)=(-2)^2+4(-2)+1=-3 q ( − 2 ) = ( − 2 ) 2 + 4 ( − 2 ) + 1 = − 3 .
Flashcard 13: Find s ( 3 ) s(3) s ( 3 ) if s ( x ) = x + 1 2 s(x)=\frac{x+1}{2} s ( x ) = 2 x + 1 . Answer: s ( 3 ) = 2 s(3)=2 s ( 3 ) = 2 . Substitute x = 3 x=3 x = 3 : s ( 3 ) = 3 + 1 2 = 2 s(3)=\frac{3+1}{2}=2 s ( 3 ) = 2 3 + 1 = 2 .
Flashcard 14: Find g ( x + 2 ) g(x+2) g ( x + 2 ) if g ( x ) = 3 x − 1 g(x)=3x-1 g ( x ) = 3 x − 1 . Answer: g ( x + 2 ) = 3 x + 5 g(x+2)=3x+5 g ( x + 2 ) = 3 x + 5 . Replace x x x with ( x + 2 ) (x+2) ( x + 2 ) : g ( x + 2 ) = 3 ( x + 2 ) − 1 = 3 x + 5 g(x+2)=3(x+2)-1=3x+5 g ( x + 2 ) = 3 ( x + 2 ) − 1 = 3 x + 5 .
Flashcard 15: What does the statement f ( x ) < 0 f(x)<0 f ( x ) < 0 describe? Answer: Inputs x x x where the function output is negative. The function is below the x x x -axis at these inputs.
Flashcard 16: Find f ( − 3 ) f(-3) f ( − 3 ) if f ( x ) = x 2 − 1 f(x)=x^2-1 f ( x ) = x 2 − 1 . Answer: f ( − 3 ) = 8 f(-3)=8 f ( − 3 ) = 8 . Substitute x = − 3 x=-3 x = − 3 : f ( − 3 ) = ( − 3 ) 2 − 1 = 8 f(-3)=(-3)^2-1=8 f ( − 3 ) = ( − 3 ) 2 − 1 = 8 .
Flashcard 17: Find p ( 1 ) p(1) p ( 1 ) if p ( x ) = 3 x 2 + 2 x p(x)=3x^2+2x p ( x ) = 3 x 2 + 2 x . Answer: p ( 1 ) = 5 p(1)=5 p ( 1 ) = 5 . Substitute x = 1 x=1 x = 1 : p ( 1 ) = 3 ( 1 ) 2 + 2 ( 1 ) = 5 p(1)=3(1)^2+2(1)=5 p ( 1 ) = 3 ( 1 ) 2 + 2 ( 1 ) = 5 .
Flashcard 18: Find r ( 5 ) r(5) r ( 5 ) if r ( x ) = 10 − 2 x r(x)=10-2x r ( x ) = 10 − 2 x . Answer: r ( 5 ) = 0 r(5)=0 r ( 5 ) = 0 . Substitute x = 5 x=5 x = 5 : r ( 5 ) = 10 − 2 ( 5 ) = 0 r(5)=10-2(5)=0 r ( 5 ) = 10 − 2 ( 5 ) = 0 .
Flashcard 19: What does the statement f ( x ) > 0 f(x)>0 f ( x ) > 0 describe? Answer: Inputs x x x where the function output is positive. The function is above the x x x -axis at these inputs.
Flashcard 20: Find h ( 4 ) h(4) h ( 4 ) if h ( x ) = 2 ( x − 1 ) h(x)=2(x-1) h ( x ) = 2 ( x − 1 ) . Answer: h ( 4 ) = 6 h(4)=6 h ( 4 ) = 6 . Substitute x = 4 x=4 x = 4 : h ( 4 ) = 2 ( 4 − 1 ) = 6 h(4)=2(4-1)=6 h ( 4 ) = 2 ( 4 − 1 ) = 6 .
Flashcard 21: Interpret C ( 10 ) = 25 C(10)=25 C ( 10 ) = 25 if C ( d ) C(d) C ( d ) is the cost in dollars for d d d miles. Answer: A 10 10 10 -mile trip costs 25 25 25 dollars. Function notation describes real-world relationships.
Flashcard 22: What does 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 . Shows the change in function values between inputs.
Flashcard 23: What is the difference between f ( x ) f(x) f ( x ) and f ⋅ x f\cdot x f ⋅ x ? Answer: f ( x ) f(x) f ( x ) is a function value; f ⋅ x f\cdot x f ⋅ x is multiplication. Parentheses indicate function evaluation, not multiplication.
Flashcard 24: Find g ( x + 2 ) g(x+2) g ( x + 2 ) if g ( x ) = 3 x − 1 g(x)=3x-1 g ( x ) = 3 x − 1 . Answer: g ( x + 2 ) = 3 x + 5 g(x+2)=3x+5 g ( x + 2 ) = 3 x + 5 . Replace x x x with ( x + 2 ) (x+2) ( x + 2 ) : g ( x + 2 ) = 3 ( x + 2 ) − 1 = 3 x + 5 g(x+2)=3(x+2)-1=3x+5 g ( x + 2 ) = 3 ( x + 2 ) − 1 = 3 x + 5 .
Flashcard 25: What is the range of a function in words? Answer: The set of all possible output values. All values the function can produce as outputs.
Flashcard 26: Find t ( − 1 ) t(-1) t ( − 1 ) if t ( x ) = ∣ x ∣ + 2 t(x)=\lvert x\rvert+2 t ( x ) = ∣ x ∣ + 2 . Answer: t ( − 1 ) = 3 t(-1)=3 t ( − 1 ) = 3 . Substitute x = − 1 x=-1 x = − 1 : t ( − 1 ) = ∣ − 1 ∣ + 2 = 3 t(-1)=\lvert-1\rvert+2=3 t ( − 1 ) = ∣ − 1 ∣ + 2 = 3 .
Flashcard 27: Find f ( 2 ) f(2) f ( 2 ) if the table gives f ( 2 ) = 9 f(2)=9 f ( 2 ) = 9 . Answer: f ( 2 ) = 9 f(2)=9 f ( 2 ) = 9 . Read function value directly from table.
Flashcard 28: Find the domain of f ( x ) = x + 4 f(x)=\sqrt{x+4} f ( x ) = x + 4 in inequality form. Answer: x ≥ − 4 x\ge -4 x ≥ − 4 . Square root requires x + 4 ≥ 0 x+4\ge 0 x + 4 ≥ 0 .
Flashcard 29: Find h ( 2 x ) h(2x) h ( 2 x ) if h ( x ) = x 2 − 4 h(x)=x^2-4 h ( x ) = x 2 − 4 . Answer: h ( 2 x ) = 4 x 2 − 4 h(2x)=4x^2-4 h ( 2 x ) = 4 x 2 − 4 . Replace x x x with 2 x 2x 2 x : h ( 2 x ) = ( 2 x ) 2 − 4 = 4 x 2 − 4 h(2x)=(2x)^2-4=4x^2-4 h ( 2 x ) = ( 2 x ) 2 − 4 = 4 x 2 − 4 .
Flashcard 30: Find q ( 1 ) q(1) q ( 1 ) if q ( x ) = 4 x + 1 q(x)=\frac{4}{x+1} q ( x ) = x + 1 4 . Answer: q ( 1 ) = 2 q(1)=2 q ( 1 ) = 2 . Substitute x = 1 x=1 x = 1 : q ( 1 ) = 4 1 + 1 = 2 q(1)=\frac{4}{1+1}=2 q ( 1 ) = 1 + 1 4 = 2 .
Flashcard 31: Find g ( 2 ) g(2) g ( 2 ) if g ( x ) = 1 2 x − 4 g(x)=\frac{1}{2}x-4 g ( x ) = 2 1 x − 4 . Answer: g ( 2 ) = − 3 g(2)=-3 g ( 2 ) = − 3 . Substitute x = 2 x=2 x = 2 : g ( 2 ) = 1 2 ( 2 ) − 4 = − 3 g(2)=\frac{1}{2}(2)-4=-3 g ( 2 ) = 2 1 ( 2 ) − 4 = − 3 .
Flashcard 32: What is the domain of f ( x ) = 1 x + 3 f(x)=\frac{1}{x+3} f ( x ) = x + 3 1 in set notation? Answer: All real x x x such that x ≠ − 3 x\neq -3 x = − 3 . Denominator cannot equal zero, so exclude x = − 3 x=-3 x = − 3 .
Flashcard 33: What is the meaning of the statement f ( 3 ) = 7 f(3)=7 f ( 3 ) = 7 ? Answer: When x = 3 x=3 x = 3 , the function output is 7 7 7 . Substitute x = 3 x=3 x = 3 into the function to get output 7 7 7 .
Flashcard 34: What does f ( 0 ) f(0) f ( 0 ) represent on the graph of y = f ( x ) y=f(x) y = f ( x ) ? Answer: The y y y -intercept value. When x = 0 x=0 x = 0 , output gives y y y -intercept.
Flashcard 35: Identify whether f ( − 1 ) f(-1) f ( − 1 ) is defined if f ( x ) = x + 1 f(x)=\sqrt{x+1} f ( x ) = x + 1 . Answer: f ( − 1 ) f(-1) f ( − 1 ) is defined and equals 0 0 0 . Square root of 0 0 0 is defined: f ( − 1 ) = 0 = 0 f(-1)=\sqrt{0}=0 f ( − 1 ) = 0 = 0 .
Flashcard 36: Identify the input and output in the ordered pair ( x , f ( x ) ) (x,f(x)) ( x , f ( x )) . Answer: Input is x x x ; output is f ( x ) f(x) f ( x ) . First coordinate is input, second is output.
Flashcard 37: Find u ( − 4 ) u(-4) u ( − 4 ) if u ( x ) = ∣ x + 1 ∣ u(x)=\lvert x+1\rvert u ( x ) = ∣ x + 1 ∣ . Answer: u ( − 4 ) = 3 u(-4)=3 u ( − 4 ) = 3 . Substitute x = − 4 x=-4 x = − 4 : u ( − 4 ) = ∣ − 4 + 1 ∣ = 3 u(-4)=\lvert-4+1\rvert=3 u ( − 4 ) = ∣ − 4 + 1 ∣ = 3 .
Flashcard 38: What is the domain of f ( x ) = 2 − x f(x)=\sqrt{2-x} f ( x ) = 2 − x in inequality form? Answer: x ≤ 2 x\le 2 x ≤ 2 . Square root requires non-negative radicand.
Flashcard 39: Find h ( 0 ) h(0) h ( 0 ) if h ( x ) = 5 − 3 x h(x)=5-3x h ( x ) = 5 − 3 x . Answer: h ( 0 ) = 5 h(0)=5 h ( 0 ) = 5 . Substitute x = 0 x=0 x = 0 : h ( 0 ) = 5 − 3 ( 0 ) = 5 h(0)=5-3(0)=5 h ( 0 ) = 5 − 3 ( 0 ) = 5 .
Flashcard 40: What is the domain of f ( x ) = 5 x 2 − 9 f(x)=\frac{5}{x^2-9} f ( x ) = x 2 − 9 5 in restriction form? Answer: All real x x x such that x ≠ 3 x\neq 3 x = 3 and x ≠ − 3 x\neq -3 x = − 3 . Factor: x 2 − 9 = ( x − 3 ) ( x + 3 ) x^2-9=(x-3)(x+3) x 2 − 9 = ( x − 3 ) ( x + 3 ) , exclude both zeros.
Flashcard 41: What point corresponds to f ( a ) = b f(a)=b f ( a ) = b on the graph of y = f ( x ) y=f(x) y = f ( x ) ? Answer: The point ( a , b ) (a,b) ( a , b ) . Function notation corresponds to coordinate pairs.
Flashcard 42: What is the meaning of the statement f ( a ) = b f(a)=b f ( a ) = b ? Answer: When x = a x=a x = a , the function output is b b b . General form showing input a a a produces output b b b .
Flashcard 43: Find f ( 4 ) f(4) f ( 4 ) if f ( x ) = 2 x + 5 f(x)=2x+5 f ( x ) = 2 x + 5 . Answer: f ( 4 ) = 13 f(4)=13 f ( 4 ) = 13 . Substitute x = 4 x=4 x = 4 : f ( 4 ) = 2 ( 4 ) + 5 = 13 f(4)=2(4)+5=13 f ( 4 ) = 2 ( 4 ) + 5 = 13 .
Flashcard 44: Find the domain of f ( x ) = x f(x)=\sqrt{x} f ( x ) = x in inequality form. Answer: x ≥ 0 x\ge 0 x ≥ 0 . Square root requires non-negative radicand.
Flashcard 45: Find p ( 0 ) p(0) p ( 0 ) if p ( x ) = x 3 − 2 x p(x)=x^3-2x p ( x ) = x 3 − 2 x . Answer: p ( 0 ) = 0 p(0)=0 p ( 0 ) = 0 . Substitute x = 0 x=0 x = 0 : p ( 0 ) = 0 3 − 2 ( 0 ) = 0 p(0)=0^3-2(0)=0 p ( 0 ) = 0 3 − 2 ( 0 ) = 0 .
Flashcard 46: Find p ( x − 1 ) p(x-1) p ( x − 1 ) if p ( x ) = x 2 p(x)=x^2 p ( x ) = x 2 . Answer: p ( x − 1 ) = ( x − 1 ) 2 p(x-1)=(x-1)^2 p ( x − 1 ) = ( x − 1 ) 2 . Replace x x x with ( x − 1 ) (x-1) ( x − 1 ) : p ( x − 1 ) = ( x − 1 ) 2 p(x-1)=(x-1)^2 p ( x − 1 ) = ( x − 1 ) 2 .
Flashcard 47: Find g ( 6 ) g(6) g ( 6 ) if g ( x ) = x 3 + 1 g(x)=\frac{x}{3}+1 g ( x ) = 3 x + 1 . Answer: g ( 6 ) = 3 g(6)=3 g ( 6 ) = 3 . Substitute x = 6 x=6 x = 6 : g ( 6 ) = 6 3 + 1 = 3 g(6)=\frac{6}{3}+1=3 g ( 6 ) = 3 6 + 1 = 3 .
Flashcard 48: Find the domain of f ( x ) = x + 1 x − 7 f(x)=\frac{x+1}{x-7} f ( x ) = x − 7 x + 1 in restriction form. Answer: All real x x x such that x ≠ 7 x\neq 7 x = 7 . Denominator cannot equal zero.
Flashcard 49: What does f ( x + h ) f(x+h) f ( x + h ) mean in function notation? Answer: The output when the input is x + h x+h x + h . Substitute the entire expression ( x + h ) (x+h) ( x + h ) for x x x .
Flashcard 50: Interpret T ( 3 ) = 68 T(3)=68 T ( 3 ) = 68 if T ( h ) T(h) T ( h ) is temperature after h h h hours. Answer: After 3 3 3 hours, the temperature is 68 68 68 degrees. Function value represents temperature at specific time.
Flashcard 51: Find p ( 0 ) p(0) p ( 0 ) if p ( x ) = x 3 − 2 x p(x)=x^3-2x p ( x ) = x 3 − 2 x . Answer: p ( 0 ) = 0 p(0)=0 p ( 0 ) = 0 . Substitute x = 0 x=0 x = 0 : p ( 0 ) = 0 3 − 2 ( 0 ) = 0 p(0)=0^3-2(0)=0 p ( 0 ) = 0 3 − 2 ( 0 ) = 0 .
Flashcard 52: Find t ( − 1 ) t(-1) t ( − 1 ) if t ( x ) = ∣ x ∣ + 2 t(x)=\lvert x\rvert+2 t ( x ) = ∣ x ∣ + 2 . Answer: t ( − 1 ) = 3 t(-1)=3 t ( − 1 ) = 3 . Substitute x = − 1 x=-1 x = − 1 : t ( − 1 ) = ∣ − 1 ∣ + 2 = 3 t(-1)=\lvert-1\rvert+2=3 t ( − 1 ) = ∣ − 1 ∣ + 2 = 3 .
Flashcard 53: Find the domain of f ( x ) = 1 ( x − 1 ) ( x + 2 ) f(x)=\frac{1}{(x-1)(x+2)} f ( x ) = ( x − 1 ) ( x + 2 ) 1 in restriction form. Answer: All real x x x such that x ≠ 1 x\neq 1 x = 1 and x ≠ − 2 x\neq -2 x = − 2 . Both factors in denominator cannot equal zero.
Flashcard 54: Find f ( a ) f(a) f ( a ) if f ( x ) = x 2 + 3 f(x)=x^2+3 f ( x ) = x 2 + 3 . Answer: f ( a ) = a 2 + 3 f(a)=a^2+3 f ( a ) = a 2 + 3 . Replace x x x with a a a in the function expression.
Flashcard 55: What is the standard notation for the output variable of f ( x ) f(x) f ( x ) ? Answer: y = f ( x ) y=f(x) y = f ( x ) . Standard way to represent function output.
Flashcard 56: Interpret d ( 2.5 ) = 150 d(2.5)=150 d ( 2.5 ) = 150 if d ( t ) d(t) d ( t ) is distance in miles after t t t hours. Answer: After 2.5 2.5 2.5 hours, the distance traveled is 150 150 150 miles. Function describes distance as function of time.
Flashcard 57: What does the notation f ( x ) f(x) f ( x ) represent in function notation? Answer: f ( x ) f(x) f ( x ) is the output value of function f f f for input x x x . Function notation shows the input-output relationship.
Flashcard 58: Find f ( a ) f(a) f ( a ) if f ( x ) = x 2 + 3 f(x)=x^2+3 f ( x ) = x 2 + 3 . Answer: f ( a ) = a 2 + 3 f(a)=a^2+3 f ( a ) = a 2 + 3 . Replace x x x with a a a in the function expression.
Flashcard 59: Identify x x x if f ( x ) = 12 f(x)=12 f ( x ) = 12 and the table shows f ( 4 ) = 12 f(4)=12 f ( 4 ) = 12 . Answer: x = 4 x=4 x = 4 . Find input that produces given output.
Flashcard 60: Find m ( 3 ) m(3) m ( 3 ) if m ( x ) = x + 1 m(x)=\sqrt{x+1} m ( x ) = x + 1 . Answer: m ( 3 ) = 2 m(3)=2 m ( 3 ) = 2 . Substitute x = 3 x=3 x = 3 : m ( 3 ) = 3 + 1 = 2 m(3)=\sqrt{3+1}=2 m ( 3 ) = 3 + 1 = 2 .