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

Answer:

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.

Step-by-step explanation:

Answer 2

Answer:

$x\geq0$   if   $f(x)=\sqrt{x}-3$

$x\geq3$   if   $f(x)=\sqrt{x-3}$

Step-by-step explanation:

If  $f(x)=\sqrt{x}-3$ 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

$(f*g)(x) = f(x)*g(x)\\\\(f*g)(x) =(\sqrt{x}-3)*(1-x^2)$

$(f*g)(x) =\sqrt{x}-x^2(\sqrt{x})-3+3x^2$

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) = $\sqrt{x-3}$ then the domain of f(x) is

$x-3\geq0\\\\x\geq3$

Therefore

$(f*g)(x) = f(x)*g(x)\\\\(f*g)(x) =(\sqrt{x-3})*(1-x^2)$

$(f*g)(x) =\sqrt{x-3}-x^2(\sqrt{x-3})$

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

$x\geq3$ or x ∈ [3, ∞)