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Question Find the range of f(x)=(x+3)2−5f(x)=(x+3)^2-5 when the domain is {−3,0,3}\{-3,0,3\}.

Studdy Solution

STEP 1

Assumptions
1. The function given is f(x)=(x+3)2−5f(x)=(x+3)^{2}-5.
2. The domain of the function is the set {−3,0,3}\{-3,0,3\}.
3. The range of a function is the set of all possible output values (function values), which result from using the function formula.

STEP 2

Calculate the value of the function when x=−3x=-3.
f(−3)=(−3+3)2−5f(-3)=(-3+3)^{2}-5

STEP 3

Simplify the expression inside the parentheses.
f(−3)=(0)2−5f(-3)=(0)^{2}-5

STEP 4

Calculate the square of 0.
f(−3)=0−5f(-3)=0-5

STEP 5

Subtract 5 from 0.
f(−3)=−5f(-3)=-5

STEP 6

Calculate the value of the function when x=0x=0.
f(0)=(0+3)2−5f(0)=(0+3)^{2}-5

STEP 7

Simplify the expression inside the parentheses.
f(0)=(3)2−5f(0)=(3)^{2}-5

STEP 8

Calculate the square of 3.
f(0)=9−5f(0)=9-5

STEP 9

Subtract 5 from 9.
f(0)=4f(0)=4

STEP 10

Calculate the value of the function when x=3x=3.
f(3)=(3+3)2−5f(3)=(3+3)^{2}-5

STEP 11

Simplify the expression inside the parentheses.
f(3)=(6)2−5f(3)=(6)^{2}-5

STEP 12

Calculate the square of 6.
f(3)=36−5f(3)=36-5

STEP 13

Subtract 5 from 36.
f(3)=31f(3)=31

STEP 14

Now that we have calculated the function values for all elements of the domain, we can determine the range. The range is the set of all function values we have found.
Range={f(−3),f(0),f(3)}Range = \{f(-3), f(0), f(3)\}

STEP 15

Substitute the function values into the set to define the range.
Range={−5,4,31}Range = \{-5, 4, 31\}
The range of the function f(x)=(x+3)2−5f(x)=(x+3)^{2}-5 when the domain is {−3,0,3}\{-3,0,3\} is {−5,4,31}\{-5, 4, 31\}.

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