Question

If f(x)= sqrtx-3 and g(x)=1-x^2, then what do you notice about the domain of (fxg)(x)?

Answer 1

The domain of f, and thus the range of g, is restricted to values greater than or equal to 3.

If 1 minus x squared is greater than or equal to 3, then x squared must be less than –2.

Since x squared cannot be less than a negative number, the function is undefined for all values of x.

Explanation:

Answer 2

`x\geq0` if
`f(x)=√(x)-3`

`x\geq3` if
`f(x)=√(x-3)`

Explanation:

If
`f(x)=√(x)-3` then the domain of f(x) is all positive real numbers. This is
`x\geq0`

On the other hand the domain of g(x) would be all real numbers because it is a polynomial function

Therefore

`(f*g)(x) = f(x)*g(x)\\\\(f*g)(x) =(√(x)-3)*(1-x^2)`

`(f*g)(x) =√(x)-x^2(√(x))-3+3x^2`

Then the domain of
`f(x) *g(x)` will be the same domain of
`f(x)`

All positive real numbers,
`x\geq0` or x ∈ [0, ∞)

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If f(x) =
`√(x-3)` then the domain of f(x) is

`x-3\geq0\\\\x\geq3`

Therefore

`(f*g)(x) = f(x)*g(x)\\\\(f*g)(x) =(√(x-3))*(1-x^2)`

`(f*g)(x) =√(x-3)-x^2(√(x-3))`

Then the domain of
`f(x) *g(x)` will be the same domain of
`f(x)`

`x\geq3` or x ∈ [3, ∞)