Solved on Sep 12, 2023

Find composite functions and their domains for $f(x)=\sqrt{x}$ and $g(x)=8x+9$ . (a) $f \circ g$ : $\sqrt{8x+9}$ , domain $x \geq -\frac{9}{8}$ . (b) $g \circ f$ : $8\sqrt{x}+9$ .

STEP 1

Assumptions1. The functions are $f(x)=\sqrt{x}$ and $g(x)=8x+9$ . We need to find the composite functions $f \circ g$ , $g \circ f$ , $f \circ f$ , and $g \circ g$
3. We also need to state the domain of each composite function

STEP 2

First, let's find the composite function $f \circ g$ . This is done by replacing every instance of $x$ in $f(x)$ with $g(x)$ .
$(f \circ g)(x) = f(g(x))$

STEP 3

Now, plug in the given functions $f(x)$ and $g(x)$ to find $f \circ g$ .
$(f \circ g)(x) = f(8x+9)$

STEP 4

Substitute $8x+9$ into $f(x)$ .
$(f \circ g)(x) = \sqrt{8x+9}$

STEP 5

Next, let's find the domain of $f \circ g$ . The domain of a function is the set of all possible input values (x-values) which will produce a valid output. For a square root function, the value inside the square root (the radicand) must be greater than or equal to0.
$8x+9 \geq0$

STEP 6

olve the inequality to find the domain of $f \circ g$ .
$x \geq -\frac{9}{8}$ So, the domain of $f \circ g$ is $\left\{x \mid x \geq-\frac{9}{8}\right\}$ .

STEP 7

Now, let's find the composite function $g \circ f$ . This is done by replacing every instance of $x$ in $g(x)$ with $f(x)$ .
$(g \circ f)(x) = g(f(x))$

STEP 8

Now, plug in the given functions $f(x)$ and $g(x)$ to find $g \circ f$ .
$(g \circ f)(x) = g(\sqrt{x})$

STEP 9

Substitute $\sqrt{x}$ into $g(x)$ .
$(g \circ f)(x) =8\sqrt{x}+9$

STEP 10

Next, let's find the domain of $g \circ f$ . The domain of a function is the set of all possible input values (x-values) which will produce a valid output. For a square root function, the value inside the square root (the radicand) must be greater than or equal to0.
$x \geq0$ So, the domain of $g \circ f$ is $\left\{x \mid x \geq0\right\}$ .

STEP 11

Now, let's find the composite function $f \circ f$ . This is done by replacing every instance of $x$ in $f(x)$ with $f(x)$ .
$(f \circ f)(x) = f(f(x))$

STEP 12

Now, plug in the given function $f(x)$ to find $f \circ f$ .
$(f \circ f)(x) = f(\sqrt{x})$

STEP 13

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

STEP 14

Next, let's find the domain of $f \circ f$ . The domain of a function is the set of all possible input values (x-values) which will produce a valid output. For a square root function, the value inside the square root (the radicand) must be greater than or equal to0.
$x \geq0$ So, the domain of $f \circ f$ is $\left\{x \mid x \geq0\right\}$ .

STEP 15

Finally, let's find the composite function $g \circ g$ . This is done by replacing every instance of $x$ in $g(x)$ with $g(x)$ .
$(g \circ g)(x) = g(g(x))$

STEP 16

Now, plug in the given function $g(x)$ to find $g \circ g$ .
$(g \circ g)(x) = g(8x+9)$

STEP 17

Substitute $x+9$ into $g(x)$ .
$(g \circ g)(x) =(x+9)+9$

STEP 18

implify the expression to find $g \circ g$ .
$(g \circ g)(x) =64x+72+ =64x+81$

STEP 19

The domain of $g \circ g$ is all real numbers since the function $g \circ g$ is a linear function and there are no restrictions on the input values.
So, the solutions are(a) $(f \circ g)(x)=\sqrt{8x+9}$ with domain $\left\{x \mid x \geq-\frac{9}{8}\right\}$ (b) $(g \circ f)(x)=8\sqrt{x}+9$ with domain $\left\{x \mid x \geq\right\}$ (c) $(f \circ f)(x)=\sqrt{\sqrt{x}}$ with domain $\left\{x \mid x \geq\right\}$ (d) $(g \circ g)(x)=64x+81$ with domain all real numbers.

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Find composite functions and their domains for $f(x)=\sqrt{x}$ and $g(x)=8x+9$ . (a) $f \circ g$ : $\sqrt{8x+9}$ , domain $x \geq -\frac{9}{8}$ . (b) $g \circ f$ : $8\sqrt{x}+9$ .

STEP 1

Assumptions1. The functions are $f(x)=\sqrt{x}$ and $g(x)=8x+9$ . We need to find the composite functions $f \circ g$ , $g \circ f$ , $f \circ f$ , and $g \circ g$
3. We also need to state the domain of each composite function

STEP 2

First, let's find the composite function $f \circ g$ . This is done by replacing every instance of $x$ in $f(x)$ with $g(x)$ .
$(f \circ g)(x) = f(g(x))$

STEP 3

Now, plug in the given functions $f(x)$ and $g(x)$ to find $f \circ g$ .
$(f \circ g)(x) = f(8x+9)$

STEP 4

Substitute $8x+9$ into $f(x)$ .
$(f \circ g)(x) = \sqrt{8x+9}$

STEP 5

Next, let's find the domain of $f \circ g$ . The domain of a function is the set of all possible input values (x-values) which will produce a valid output. For a square root function, the value inside the square root (the radicand) must be greater than or equal to0.
$8x+9 \geq0$

STEP 6

olve the inequality to find the domain of $f \circ g$ .
$x \geq -\frac{9}{8}$ So, the domain of $f \circ g$ is $\left\{x \mid x \geq-\frac{9}{8}\right\}$ .

STEP 7

Now, let's find the composite function $g \circ f$ . This is done by replacing every instance of $x$ in $g(x)$ with $f(x)$ .
$(g \circ f)(x) = g(f(x))$

STEP 8

Now, plug in the given functions $f(x)$ and $g(x)$ to find $g \circ f$ .
$(g \circ f)(x) = g(\sqrt{x})$

STEP 9

Substitute $\sqrt{x}$ into $g(x)$ .
$(g \circ f)(x) =8\sqrt{x}+9$

STEP 10

STEP 11

Now, let's find the composite function $f \circ f$ . This is done by replacing every instance of $x$ in $f(x)$ with $f(x)$ .
$(f \circ f)(x) = f(f(x))$

STEP 12

Now, plug in the given function $f(x)$ to find $f \circ f$ .
$(f \circ f)(x) = f(\sqrt{x})$

STEP 13

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

STEP 14

STEP 15

Finally, let's find the composite function $g \circ g$ . This is done by replacing every instance of $x$ in $g(x)$ with $g(x)$ .
$(g \circ g)(x) = g(g(x))$

STEP 16

Now, plug in the given function $g(x)$ to find $g \circ g$ .
$(g \circ g)(x) = g(8x+9)$

STEP 17

Substitute $x+9$ into $g(x)$ .
$(g \circ g)(x) =(x+9)+9$

STEP 18

implify the expression to find $g \circ g$ .
$(g \circ g)(x) =64x+72+ =64x+81$

STEP 19

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Find composite functions and their domains for f(x)=xf(x)=\sqrt{x}f(x)=x​ and g(x)=8x+9g(x)=8x+9g(x)=8x+9. (a) f∘gf \circ gf∘g: 8x+9\sqrt{8x+9}8x+9​, domain x≥−98x \geq -\frac{9}{8}x≥−89​. (b) g∘fg \circ fg∘f: 8x+98\sqrt{x}+98x​+9.

STEP 1

Assumptions1. The functions are f(x)=xf(x)=\sqrt{x}f(x)=x​ and g(x)=8x+9g(x)=8x+9g(x)=8x+9 . We need to find the composite functions f∘gf \circ gf∘g, g∘fg \circ fg∘f, f∘ff \circ ff∘f, and g∘gg \circ gg∘g3. We also need to state the domain of each composite function

STEP 2

First, let's find the composite function f∘gf \circ gf∘g. This is done by replacing every instance of xxx in f(x)f(x)f(x) with g(x)g(x)g(x).(f∘g)(x)=f(g(x))(f \circ g)(x) = f(g(x))(f∘g)(x)=f(g(x))

STEP 3

Now, plug in the given functions f(x)f(x)f(x) and g(x)g(x)g(x) to find f∘gf \circ gf∘g.(f∘g)(x)=f(8x+9)(f \circ g)(x) = f(8x+9)(f∘g)(x)=f(8x+9)

STEP 4

Substitute 8x+98x+98x+9 into f(x)f(x)f(x).(f∘g)(x)=8x+9(f \circ g)(x) = \sqrt{8x+9}(f∘g)(x)=8x+9​

STEP 5

Next, let's find the domain of f∘gf \circ gf∘g. The domain of a function is the set of all possible input values (x-values) which will produce a valid output. For a square root function, the value inside the square root (the radicand) must be greater than or equal to0.8x+9≥08x+9 \geq08x+9≥0

STEP 6

olve the inequality to find the domain of f∘gf \circ gf∘g.x≥−98x \geq -\frac{9}{8}x≥−89​So, the domain of f∘gf \circ gf∘g is {x∣x≥−98}\left\{x \mid x \geq-\frac{9}{8}\right\}{x∣x≥−89​}.

STEP 7

Now, let's find the composite function g∘fg \circ fg∘f. This is done by replacing every instance of xxx in g(x)g(x)g(x) with f(x)f(x)f(x).(g∘f)(x)=g(f(x))(g \circ f)(x) = g(f(x))(g∘f)(x)=g(f(x))

STEP 8

Now, plug in the given functions f(x)f(x)f(x) and g(x)g(x)g(x) to find g∘fg \circ fg∘f.(g∘f)(x)=g(x)(g \circ f)(x) = g(\sqrt{x})(g∘f)(x)=g(x​)

STEP 9

Substitute x\sqrt{x}x​ into g(x)g(x)g(x).(g∘f)(x)=8x+9(g \circ f)(x) =8\sqrt{x}+9(g∘f)(x)=8x​+9

STEP 10

STEP 11

Now, let's find the composite function f∘ff \circ ff∘f. This is done by replacing every instance of xxx in f(x)f(x)f(x) with f(x)f(x)f(x).(f∘f)(x)=f(f(x))(f \circ f)(x) = f(f(x))(f∘f)(x)=f(f(x))

STEP 12

Now, plug in the given function f(x)f(x)f(x) to find f∘ff \circ ff∘f.(f∘f)(x)=f(x)(f \circ f)(x) = f(\sqrt{x})(f∘f)(x)=f(x​)

STEP 13

Substitute x\sqrt{x}x​ into f(x)f(x)f(x).(f∘f)(x)=x(f \circ f)(x) = \sqrt{\sqrt{x}}(f∘f)(x)=x​​

STEP 14

STEP 15

Finally, let's find the composite function g∘gg \circ gg∘g. This is done by replacing every instance of xxx in g(x)g(x)g(x) with g(x)g(x)g(x).(g∘g)(x)=g(g(x))(g \circ g)(x) = g(g(x))(g∘g)(x)=g(g(x))

STEP 16

Now, plug in the given function g(x)g(x)g(x) to find g∘gg \circ gg∘g.(g∘g)(x)=g(8x+9)(g \circ g)(x) = g(8x+9)(g∘g)(x)=g(8x+9)

STEP 17

Substitute x+9x+9x+9 into g(x)g(x)g(x).(g∘g)(x)=(x+9)+9(g \circ g)(x) =(x+9)+9(g∘g)(x)=(x+9)+9

STEP 18

implify the expression to find g∘gg \circ gg∘g.(g∘g)(x)=64x+72+=64x+81(g \circ g)(x) =64x+72+ =64x+81(g∘g)(x)=64x+72+=64x+81

STEP 19

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Find composite functions and their domains for $f(x)=\sqrt{x}$ and $g(x)=8x+9$ . (a) $f \circ g$ : $\sqrt{8x+9}$ , domain $x \geq -\frac{9}{8}$ . (b) $g \circ f$ : $8\sqrt{x}+9$ .

Assumptions1. The functions are $f(x)=\sqrt{x}$ and $g(x)=8x+9$ . We need to find the composite functions $f \circ g$ , $g \circ f$ , $f \circ f$ , and $g \circ g$
3. We also need to state the domain of each composite function

First, let's find the composite function $f \circ g$ . This is done by replacing every instance of $x$ in $f(x)$ with $g(x)$ .
$(f \circ g)(x) = f(g(x))$

Now, plug in the given functions $f(x)$ and $g(x)$ to find $f \circ g$ .
$(f \circ g)(x) = f(8x+9)$

Substitute $8x+9$ into $f(x)$ .
$(f \circ g)(x) = \sqrt{8x+9}$

Next, let's find the domain of $f \circ g$ . The domain of a function is the set of all possible input values (x-values) which will produce a valid output. For a square root function, the value inside the square root (the radicand) must be greater than or equal to0.
$8x+9 \geq0$

olve the inequality to find the domain of $f \circ g$ .
$x \geq -\frac{9}{8}$ So, the domain of $f \circ g$ is $\left\{x \mid x \geq-\frac{9}{8}\right\}$ .

Now, let's find the composite function $g \circ f$ . This is done by replacing every instance of $x$ in $g(x)$ with $f(x)$ .
$(g \circ f)(x) = g(f(x))$

Now, plug in the given functions $f(x)$ and $g(x)$ to find $g \circ f$ .
$(g \circ f)(x) = g(\sqrt{x})$

Substitute $\sqrt{x}$ into $g(x)$ .
$(g \circ f)(x) =8\sqrt{x}+9$

Now, let's find the composite function $f \circ f$ . This is done by replacing every instance of $x$ in $f(x)$ with $f(x)$ .
$(f \circ f)(x) = f(f(x))$

Now, plug in the given function $f(x)$ to find $f \circ f$ .
$(f \circ f)(x) = f(\sqrt{x})$

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

Finally, let's find the composite function $g \circ g$ . This is done by replacing every instance of $x$ in $g(x)$ with $g(x)$ .
$(g \circ g)(x) = g(g(x))$

Now, plug in the given function $g(x)$ to find $g \circ g$ .
$(g \circ g)(x) = g(8x+9)$

Substitute $x+9$ into $g(x)$ .
$(g \circ g)(x) =(x+9)+9$

implify the expression to find $g \circ g$ .
$(g \circ g)(x) =64x+72+ =64x+81$