QuestionFind $f \circ g$ and $g \circ f$ for $f(x)=\sqrt{x+8}$ and $g(x)=x^{2}$ . Determine the domains in interval notation.

Studdy Solution

STEP 1

Assumptions1. The function $f(x)$ is defined as $f(x)=\sqrt{x+8}$ . . The function $g(x)$ is defined as $g(x)=x^{}$ .
3. The composite function $f \circ g$ means $f(g(x))$ .
4. The composite function $g \circ f$ means $g(f(x))$ .
5. The domain of a function is the set of all possible input values (x-values) which will produce a valid output.

STEP 2

First, let's find the composite function $f \circ g$ which is $f(g(x))$ .
$f \circ g(x) = f(g(x))$

STEP 3

Substitute $g(x)$ into $f(x)$ .
$f \circ g(x) = f(x^{2})$

STEP 4

Replace $x$ in $f(x)$ with $x^{2}$ .
$f \circ g(x) = \sqrt{x^{2}+8}$

STEP 5

Now, let's find the composite function $g \circ f$ which is $g(f(x))$ .
$g \circ f(x) = g(f(x))$

STEP 6

Substitute $f(x)$ into $g(x)$ .
$g \circ f(x) = g(\sqrt{x+8})$

STEP 7

Replace $x$ in $g(x)$ with $\sqrt{x+}$ .
$g \circ f(x) = (\sqrt{x+})^{2}$

STEP 8

implify the expression.
$g \circ f(x) = x+8$

STEP 9

Now, let's find the domain of $f(x)$ . The domain of $f(x)$ is the set of all $x$ such that $x+8 \geq$ (since the square root of a negative number is not a real number).
$x+8 \geq$

STEP 10

olve for $x$ .
$x \geq -8$ So the domain of $f(x)$ is $[-8, \infty)$ .

STEP 11

The domain of $g(x)$ is all real numbers because $x^{}$ is defined for all $x$ . So the domain of $g(x)$ is $(-\infty, \infty)$ .

STEP 12

Now, let's find the domain of the composite function $f \circ g(x)$ . The domain of $f \circ g(x)$ is the set of all $x$ such that $x^{2}+8 \geq0$ .
$x^{2}+8 \geq0$

STEP 13

Since $x^{2}$ is always non-negative and $8$ is positive, $x^{2}+8$ is always positive. So the domain of $f \circ g(x)$ is all real numbers, or $(-\infty, \infty)$ .

STEP 14

Finally, let's find the domain of the composite function $g \circ f(x)$ . The domain of $g \circ f(x)$ is the set of all $x$ such that $x+8 \geq0$ (since the square root of a negative number is not a real number).
$x+8 \geq0$

STEP 15

olve for $x$ .
$x \geq -8$ So the domain of $g \circ f(x)$ is $[-8, \infty)$ .
In conclusion, we have(a) $f \circ g(x) = \sqrt{x^{2}+8}$ with domain $(-\infty, \infty)$ (b) $g \circ f(x) = x+8$ with domain $[-8, \infty)$ The domain of $f(x)$ is $[-8, \infty)$ and the domain of $g(x)$ is $(-\infty, \infty)$ .

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