Question

For which pair of functions is (g circle f) (a) = StartAbsoluteValue a EndAbsoluteValue minus 2?

f(a) = a2 – 4 and g (a) = StartRoot a EndRoot
f (a) = one-half a minus 1 and g(a) = 2a – 2
f(a) = 5 + a2 and g (a) = StartRoot a minus 5 EndRoot minus 2
f(a) = 3 – 3a and g(a) = 4a – 5

Please explain in detail I am very confused about the whole thing thx for ur help in advance :)

Answer 1

Given:

`(g\circ f)(a)=|a|-2`

To find:

The functions f(x) and g(x).

Solution:

We know that,

`(g\circ f)(a)=g[f(a)]`

If
`f(a)=a^2-4` and
`g(a)=√(a)`, then

`(g\circ f)(a)=g[a^2-4]`

`(g\circ f)(a)=√(a^2-4)\\eq |a|-2`

Option A is incorrect.

If
`f(a)=(1)/(2)a-1` and
`g(a)=2a-2`, then

`(g\circ f)(a)=g[(1)/(2)a-1]`

`(g\circ f)(a)=2((1)/(2)a-1)-2`

`(g\circ f)(a)=a-2-2`

`(g\circ f)(a)=a-4\\eq |a|-2`

Option B is incorrect.

If
`f(a)=5+a^2` and
`g(a)=√(a-5)-2`, then

`(g\circ f)(a)=g[5+a^2]`

`(g\circ f)(a)=√(5+a^2-5)-2`

`(g\circ f)(a)=√(a^2)-2`

`(g\circ f)(a)=|a|-2`

Option C is correct.

If
`f(a)=3-3a` and
`g(a)=4a-5`, then

`(g\circ f)(a)=g[3-3a]`

`(g\circ f)(a)=4(3-3a)-5`

`(g\circ f)(a)=12-12a-5`

`(g\circ f)(a)=7-12x\\eq |a|-2`

Option D is incorrect.

Therefore, the correct option is C.