For the function (fog), where f(x)=√x and g(x) = 3x + 5 the restriction on the domain is: O a) x ≥-3/ b) x ≥-³/ 0020 c) there is no restriction on the domain d) x > 20/00
For the function (fog), where f(x)=√x and g(x) = 3x + 5 the restriction on the domain is: O a) x ≥-3/ b) x ≥-³/ 0020 c) there is no restriction on the domain d) x > 20/00
Chapter3: Functions
Section3.7: Inverse Functions
Problem 2SE: Why do we restrict the domain of the function f(x)=x2 to find the function's inverse?
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Need help with this two part question please
![For the function (fog), where ƒ(x) = √x and g(x) = 3x + 5 the restriction on
the domain is:
O a) x ≥ - 1/13
>
b) x ≥ ³/
>
c) there is no restriction on the domain
d) x>- 3/32](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Facb08dbd-ca29-4fdb-9f23-06f94eb41b9e%2Fcb036af0-acea-4916-929b-53eb7c5faab4%2Fb5x1fq_processed.png&w=3840&q=75)
Transcribed Image Text:For the function (fog), where ƒ(x) = √x and g(x) = 3x + 5 the restriction on
the domain is:
O a) x ≥ - 1/13
>
b) x ≥ ³/
>
c) there is no restriction on the domain
d) x>- 3/32
![For the function ()(x), where f(x) = 3x + 5 and g(x) = log₂, the restriction
on the domain would be:
a) x > 0
b) x = 1
c) there is no restriction on the domain
d) x > 0, x 1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Facb08dbd-ca29-4fdb-9f23-06f94eb41b9e%2Fcb036af0-acea-4916-929b-53eb7c5faab4%2F8rb0fyt_processed.png&w=3840&q=75)
Transcribed Image Text:For the function ()(x), where f(x) = 3x + 5 and g(x) = log₂, the restriction
on the domain would be:
a) x > 0
b) x = 1
c) there is no restriction on the domain
d) x > 0, x 1
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