Question

If `g(x) = x^2 - x +1` and `f(x) = sqrt(1/x-x)`, then find the range of `f(g(x))`?

Answer 1

.

Answer 2

I got a range of `y in [0,+sqrt(7/12)]`

Explanation:

Assuming we are limited to Real numbers
`f(g(x))=sqrt(1/(g(x))-g(x))` is only defined for `(1/(g(x))-g(x)) >= 0`

For `g(x)=x^2-x+1`
`color(white)("XXX")1/(g(x))-g(x) = 0`
only for the argument values `x=0` and `x=1`

Both `g(x)` and `1/(g(x))` have an axis of symmetry at `x=1/2`

`g(1/2)=1/4-1/2+1=3/4`

`1/(g(1/2))=1/(g(1/2))=4/3`

The maximum (positive) difference between `1/(g(x))` and `g(x)`
is `4/3-3/4=7/12`

Remembering that `1/g(x)-g(x)` must be `>=0`
we see that the legal range of `1/(g(x))-g(x)` must be `[0,7/12]`
and
`f(g(x))=sqrt(1/(g(x))-g(x))` must have a range of `[0,sqrt(7/12)]`