QuestionFind $f[g(x)]$ and $g[f(x)]$ for $f(x)=\frac{4}{x+3}$ and $g(x)=x^{2}-3$ .

Studdy Solution

STEP 1

Assumptions1. The function $f(x)$ is defined as $f(x)=\frac{4}{x+3}$ . The function $g(x)$ is defined as $g(x)=x^{}-3$

STEP 2

First, let's find $f[g(x)]$ . This means we are substituting $g(x)$ into the function $f(x)$ .
$f[g(x)] = f(x^{2}-)$

STEP 3

Now, plug in the function $g(x)$ into $f(x)$ .
$f[g(x)] = \frac{}{(x^{2}-3)+3}$

STEP 4

implify the expression.
$f[g(x)] = \frac{4}{x^{2}}$

STEP 5

Next, let's find $g[f(x)]$ . This means we are substituting $f(x)$ into the function $g(x)$ .
$g[f(x)] = g\left(\frac{4}{x+3}\right)$

STEP 6

Now, plug in the function $f(x)$ into $g(x)$ .
$g[f(x)] = \left(\frac{4}{x+3}\right)^{2}-3$

STEP 7

implify the expression.
$g[f(x)] = \frac{16}{(x+3)^{2}}-3$ So, $f[g(x)] = \frac{4}{x^{2}}$ and $g[f(x)] = \frac{16}{(x+3)^{2}}-3$ .

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QuestionFind $f[g(x)]$ and $g[f(x)]$ for $f(x)=\frac{4}{x+3}$ and $g(x)=x^{2}-3$ .

Studdy Solution

STEP 1

Assumptions1. The function $f(x)$ is defined as $f(x)=\frac{4}{x+3}$ . The function $g(x)$ is defined as $g(x)=x^{}-3$

STEP 2

First, let's find $f[g(x)]$ . This means we are substituting $g(x)$ into the function $f(x)$ .
$f[g(x)] = f(x^{2}-)$

STEP 3

Now, plug in the function $g(x)$ into $f(x)$ .
$f[g(x)] = \frac{}{(x^{2}-3)+3}$

STEP 4

implify the expression.
$f[g(x)] = \frac{4}{x^{2}}$

STEP 5

Next, let's find $g[f(x)]$ . This means we are substituting $f(x)$ into the function $g(x)$ .
$g[f(x)] = g\left(\frac{4}{x+3}\right)$

STEP 6

Now, plug in the function $f(x)$ into $g(x)$ .
$g[f(x)] = \left(\frac{4}{x+3}\right)^{2}-3$

STEP 7

implify the expression.
$g[f(x)] = \frac{16}{(x+3)^{2}}-3$ So, $f[g(x)] = \frac{4}{x^{2}}$ and $g[f(x)] = \frac{16}{(x+3)^{2}}-3$ .

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QuestionFind f[g(x)]f[g(x)]f[g(x)] and g[f(x)]g[f(x)]g[f(x)] for f(x)=4x+3f(x)=\frac{4}{x+3}f(x)=x+34​ and g(x)=x2−3g(x)=x^{2}-3g(x)=x2−3.

Studdy Solution

STEP 1

Assumptions1. The function f(x)f(x)f(x) is defined as f(x)=4x+3f(x)=\frac{4}{x+3}f(x)=x+34​ . The function g(x)g(x)g(x) is defined as g(x)=x−3g(x)=x^{}-3g(x)=x−3

STEP 2

First, let's find f[g(x)]f[g(x)]f[g(x)]. This means we are substituting g(x)g(x)g(x) into the function f(x)f(x)f(x).f[g(x)]=f(x2−)f[g(x)] = f(x^{2}-)f[g(x)]=f(x2−)

STEP 3

Now, plug in the function g(x)g(x)g(x) into f(x)f(x)f(x).f[g(x)]=(x2−3)+3f[g(x)] = \frac{}{(x^{2}-3)+3}f[g(x)]=(x2−3)+3​

STEP 4

implify the expression.f[g(x)]=4x2f[g(x)] = \frac{4}{x^{2}}f[g(x)]=x24​

STEP 5

Next, let's find g[f(x)]g[f(x)]g[f(x)]. This means we are substituting f(x)f(x)f(x) into the function g(x)g(x)g(x).g[f(x)]=g(4x+3)g[f(x)] = g\left(\frac{4}{x+3}\right)g[f(x)]=g(x+34​)

STEP 6

Now, plug in the function f(x)f(x)f(x) into g(x)g(x)g(x).g[f(x)]=(4x+3)2−3g[f(x)] = \left(\frac{4}{x+3}\right)^{2}-3g[f(x)]=(x+34​)2−3

STEP 7

implify the expression.g[f(x)]=16(x+3)2−3g[f(x)] = \frac{16}{(x+3)^{2}}-3g[f(x)]=(x+3)216​−3So, f[g(x)]=4x2f[g(x)] = \frac{4}{x^{2}}f[g(x)]=x24​ and g[f(x)]=16(x+3)2−3g[f(x)] = \frac{16}{(x+3)^{2}}-3g[f(x)]=(x+3)216​−3.

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QuestionFind $f[g(x)]$ and $g[f(x)]$ for $f(x)=\frac{4}{x+3}$ and $g(x)=x^{2}-3$ .

Assumptions1. The function $f(x)$ is defined as $f(x)=\frac{4}{x+3}$ . The function $g(x)$ is defined as $g(x)=x^{}-3$

First, let's find $f[g(x)]$ . This means we are substituting $g(x)$ into the function $f(x)$ .
$f[g(x)] = f(x^{2}-)$

Now, plug in the function $g(x)$ into $f(x)$ .
$f[g(x)] = \frac{}{(x^{2}-3)+3}$

implify the expression.
$f[g(x)] = \frac{4}{x^{2}}$

Next, let's find $g[f(x)]$ . This means we are substituting $f(x)$ into the function $g(x)$ .
$g[f(x)] = g\left(\frac{4}{x+3}\right)$

Now, plug in the function $f(x)$ into $g(x)$ .
$g[f(x)] = \left(\frac{4}{x+3}\right)^{2}-3$

implify the expression.
$g[f(x)] = \frac{16}{(x+3)^{2}}-3$ So, $f[g(x)] = \frac{4}{x^{2}}$ and $g[f(x)] = \frac{16}{(x+3)^{2}}-3$ .