# If f(x)=2(x)^2+5 and g(x)=sqrt(x-5) then find the domain and range of (fog)x?

`f(x) = 2x^2+5`

domain of f(x) is any integer, since x^2 is defined for every real number.

`g(x) = sqrt(x-5)`

domain of g(x) is any real number which is equal or greater than 5, `x gt=5`

(since sqrt is not defined for negative numbers).

`(fog)(x) = f(g(x))`

`(fog)(x) = f(sqrt(x-5))`

...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

`f(x) = 2x^2+5`

domain of f(x) is any integer, since x^2 is defined for every real number.

`g(x) = sqrt(x-5)`

domain of g(x) is any real number which is equal or greater than 5, `x gt=5`

(since sqrt is not defined for negative numbers).

`(fog)(x) = f(g(x))`

`(fog)(x) = f(sqrt(x-5))`

`(fog)(x) = 2(sqrt(x-5))^2+5`

`(fog)(x) = 2(x-5)+5`

`(fog)(x) = 2x-5`

Since the g(x) is not defiend at x<5 , then the domain of of (fog)(x) is equal to any real number equal or greater than 5. `x inR, x>=5`

The range of (fog)(x) can be found this,

let y be (fog)(x),

then,

`y = 2x-5`

`x = (y+5)/2`

but we know, `x gt= 5` ,

then,

`(y+5)/2 gt=5`

`(y+5)gt= 10`

`ygt=5`

The range of values for (fog)(x) is also equal or greater than 5.

So,

Domain of (fog)(x) is x in R,` x gt= 5`

Range of (fog)(x) is x in R, `x gt= 5`

Approved by eNotes Editorial Team