Match the functions to their ranges when the domain is {1, 2}. Tiles f(x) = 3x + 5 f(x) = x2 – 2x – 5 f(x) = (x + 5)x2 f(x) = 4 – x Pairs {-6, -5} {8, 11} {3, 2} {6, 28}

The first thing you should do for this case is to replace the following values in each of the functions:
x = 1
x = 2
The results will be the range for each function.
The results are:
f (x) = 3x + 5 -----------> {8, 11}
f (x) = 4 - x -------------> {3, 2}
f (x) = x2 - 2x - 5 -----> {-6, -5}
f (x) = (x + 5) x2 --------> {6, 28}

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pemanuel pemanuel · Answer 2 · 2019-09-11T16:53:27+02:00

Answer:

f(x) = 3x + 5 with {8, 11}, [tex]f(x) =x^2-2x-5[/tex] with {-6, -5}, [tex]f(x)=(x+5)x^2[/tex] with {6, 28} and f(x) = 4-x with {3, 2}.

Step-by-step explanation:

The domain are the possible x-values and the range are the y-values obtained with the domain. Then, let's calculate the range of each function with the given domain.

f(x) = 3x+5:

f(1) = 3(1)+5 = 3+5 = 8.

f(2) = 3(2)+5 = 6+5 = 11.

Then, the range of f(x) = 3x+5 is {8, 11}.

[tex]f(x) =x^2-2x-5:[/tex]

[tex]f(1) = 1^2-2(1)-5 = 1-2-5 = -6.[/tex]

[tex]f(2) = 2^2-2(2)-5 = 4-4-5 = -5.[/tex]

Then, the range of [tex]f(x) =x^2-2x-5[/tex] is {-6, -5}.

[tex]f(x)=(x+5)x^2:[/tex]

[tex]f(1)=(1+5)1^2=6*1=6.[/tex]

[tex]f(2) = (2+5)2^2=7*4 = 28.[/tex]

Then, the range of [tex]f(x)=(x+5)x^2[/tex] is {6, 28}.

f(x) = 4-x:

f(1) = 4-1=3.

f(2) = 4-2 = 2.

Then, the range od f(x) = 4-x is {3,2}.

Match the functions to their ranges when the domain is {1, 2}. Tiles f(x) = 3x + 5 f(x) = x2 – 2x – 5 f(x) = (x + 5)x2 f(x) = 4 – x Pairs {-6, ­-5} {8, 11} {3, 2} {6, 28}

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Match the functions to their ranges when the domain is {1, 2}. Tiles f(x) = 3x + 5 f(x) = x2 – 2x – 5 f(x) = (x + 5)x2 f(x) = 4 – x Pairs {-6, -5} {8, 11} {3, 2} {6, 28}