Question

1. find the composition of the function.

$f(x)=\sqrt{x}$ and $g(x)=x^{2}-6x + 9$ find $f(g(x))$.
$x^{2}+3$
$x + 3$
$x - 3$
$x^{2}-3$

1. find the composition of the function.

$f(x)=\sqrt{x}$ and $g(x)=x^{2}-6x + 9$ find $g(f(x))$.
$x+6\sqrt{x}-9$
$x - 6\sqrt{x}+9$
$7\sqrt{x}-9$
$7\sqrt{x}+9$

1. find the equation of the circle with center $(-2,5)$ and radius 5.

$(x + 2)^{2}+(y - 5)^{2}=25$
$(x + 2)^{2}-(y - 5)^{2}=25$
$(x - 5)^{2}+(y + 2)^{2}=25$
$(x - 2)^{2}+(y + 5)^{2}=25$

Explanation:

Step1: Find \(f(g(x))\)

Given \(f(x)=\sqrt{x}\) and \(g(x)=x^{2}-6x + 9\), substitute \(g(x)\) into \(f(x)\). So \(f(g(x))=\sqrt{g(x)}=\sqrt{x^{2}-6x + 9}\). Since \(x^{2}-6x + 9=(x - 3)^{2}\), then \(f(g(x))=\sqrt{(x - 3)^{2}}=\vert x-3\vert\). Assuming we are working in the domain where the result simplifies to a non - absolute value form for real - valued functions in a typical algebraic context, \(f(g(x))=x - 3\) (when \(x\geq3\)).

Step2: Find \(g(f(x))\)

Substitute \(f(x)\) into \(g(x)\). Given \(f(x)=\sqrt{x}\) and \(g(x)=x^{2}-6x + 9\), then \(g(f(x))=(\sqrt{x})^{2}-6\sqrt{x}+9=x - 6\sqrt{x}+9\).

Step3: Find the equation of the circle

The standard form of the equation of a circle is \((x - a)^{2}+(y - b)^{2}=r^{2}\), where \((a,b)\) is the center and \(r\) is the radius. Given the center \((-2,5)\) and radius \(r = 5\), we substitute \(a=-2\), \(b = 5\) and \(r = 5\) into the formula. So \((x-(-2))^{2}+(y - 5)^{2}=5^{2}\), which simplifies to \((x + 2)^{2}+(y - 5)^{2}=25\).

Answer:

1. C. \(x - 3\)
2. B. \(x - 6\sqrt{x}+9\)
3. A. \((x + 2)^{2}+(y - 5)^{2}=25\)