Solved on Nov 04, 2023

Find the inverse function $h^{-1}(x)$ of $h(x)=\frac{7x}{x-4}$ and its domain and range in interval notation.

STEP 1

Assumptions1. The function $h$ is one-to-one, which means that it has an inverse. . The function $h$ is defined as $h(x)=\frac{7x}{x-4}$ .

STEP 2

To find the inverse of a function, we first replace the function notation $h(x)$ with $y$ .
$y=\frac{7x}{x-4}$

STEP 3

Next, we switch the roles of $x$ and $y$ . This means we replace every $x$ in our equation with $y$ and every $y$ with $x$ .
$x=\frac{7y}{y-}$

STEP 4

Now, we solve this equation for $y$ to find $h^{-1}(x)$ . First, multiply both sides by $y-4$ to get rid of the denominator on the right side.
$(y-4)x=7y$

STEP 5

istribute $x$ on the left side.
$yx-4x=7y$

STEP 6

Rearrange the equation to isolate terms with $y$ on one side.
$yx-y=4x$

STEP 7

Factor out $y$ on the left side.
$y(x-7)=4x$

STEP 8

Finally, divide both sides by $x-7$ to solve for $y$ .
$y=\frac{4x}{x-7}$ So, $h^{-1}(x)=\frac{4x}{x-7}$ .

STEP 9

Now, let's find the domain of $h^{-}$ . The domain of a function is the set of all possible input values (x-values) which will produce a valid output. For this function, the denominator cannot be zero, as division by zero is undefined. So, we set the denominator equal to zero and solve for $x$ .
$x-7=$

STEP 10

olving the equation gives $x=7$ . This is the value that $x$ cannot take, so the domain of $h^{-}$ is all real numbers except $7$ . In interval notation, this is $(-\infty,7) \cup (7,\infty)$ .

STEP 11

The range of a function is the set of all possible output values (y-values). Since the function $h^{-}$ is a rational function and there is no restriction on the numerator, the range is all real numbers except for the value that makes the numerator zero.
$4x=0$

STEP 12

olving the equation gives $x=0$ . This is the value that $y$ cannot take, so the range of $h^{-}$ is all real numbers except $0$ . In interval notation, this is $(-\infty,0) \cup (0,\infty)$ .
So, the inverse function is $h^{-}(x)=\frac{4x}{x-7}$ , the domain of $h^{-}$ is $(-\infty,7) \cup (7,\infty)$ , and the range of $h^{-}$ is $(-\infty,0) \cup (0,\infty)$ .

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Find the inverse function $h^{-1}(x)$ of $h(x)=\frac{7x}{x-4}$ and its domain and range in interval notation.

STEP 1

Assumptions1. The function $h$ is one-to-one, which means that it has an inverse. . The function $h$ is defined as $h(x)=\frac{7x}{x-4}$ .

STEP 2

To find the inverse of a function, we first replace the function notation $h(x)$ with $y$ .
$y=\frac{7x}{x-4}$

STEP 3

Next, we switch the roles of $x$ and $y$ . This means we replace every $x$ in our equation with $y$ and every $y$ with $x$ .
$x=\frac{7y}{y-}$

STEP 4

Now, we solve this equation for $y$ to find $h^{-1}(x)$ . First, multiply both sides by $y-4$ to get rid of the denominator on the right side.
$(y-4)x=7y$

STEP 5

istribute $x$ on the left side.
$yx-4x=7y$

STEP 6

Rearrange the equation to isolate terms with $y$ on one side.
$yx-y=4x$

STEP 7

Factor out $y$ on the left side.
$y(x-7)=4x$

STEP 8

Finally, divide both sides by $x-7$ to solve for $y$ .
$y=\frac{4x}{x-7}$ So, $h^{-1}(x)=\frac{4x}{x-7}$ .

STEP 9

STEP 10

olving the equation gives $x=7$ . This is the value that $x$ cannot take, so the domain of $h^{-}$ is all real numbers except $7$ . In interval notation, this is $(-\infty,7) \cup (7,\infty)$ .

STEP 11

The range of a function is the set of all possible output values (y-values). Since the function $h^{-}$ is a rational function and there is no restriction on the numerator, the range is all real numbers except for the value that makes the numerator zero.
$4x=0$

STEP 12

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Find the inverse function h−1(x)h^{-1}(x)h−1(x) of h(x)=7xx−4h(x)=\frac{7x}{x-4}h(x)=x−47x​ and its domain and range in interval notation.

STEP 1

Assumptions1. The function hhh is one-to-one, which means that it has an inverse. . The function hhh is defined as h(x)=7xx−4h(x)=\frac{7x}{x-4}h(x)=x−47x​.

STEP 2

To find the inverse of a function, we first replace the function notation h(x)h(x)h(x) with yyy.y=7xx−4y=\frac{7x}{x-4}y=x−47x​

STEP 3

Next, we switch the roles of xxx and yyy. This means we replace every xxx in our equation with yyy and every yyy with xxx.x=7yy−x=\frac{7y}{y-}x=y−7y​

STEP 4

Now, we solve this equation for yyy to find h−1(x)h^{-1}(x)h−1(x). First, multiply both sides by y−4y-4y−4 to get rid of the denominator on the right side.(y−4)x=7y(y-4)x=7y(y−4)x=7y

STEP 5

istribute xxx on the left side.yx−4x=7yyx-4x=7yyx−4x=7y

STEP 6

Rearrange the equation to isolate terms with yyy on one side.yx−y=4xyx-y=4xyx−y=4x

STEP 7

Factor out yyy on the left side.y(x−7)=4xy(x-7)=4xy(x−7)=4x

STEP 8

Finally, divide both sides by x−7x-7x−7 to solve for yyy.y=4xx−7y=\frac{4x}{x-7}y=x−74x​So, h−1(x)=4xx−7h^{-1}(x)=\frac{4x}{x-7}h−1(x)=x−74x​.

STEP 9

STEP 10

olving the equation gives x=7x=7x=7. This is the value that xxx cannot take, so the domain of h−h^{-}h− is all real numbers except 777. In interval notation, this is (−∞,7)∪(7,∞)(-\infty,7) \cup (7,\infty)(−∞,7)∪(7,∞).

STEP 11

The range of a function is the set of all possible output values (y-values). Since the function h−h^{-}h− is a rational function and there is no restriction on the numerator, the range is all real numbers except for the value that makes the numerator zero.4x=04x=04x=0

STEP 12

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Find the inverse function $h^{-1}(x)$ of $h(x)=\frac{7x}{x-4}$ and its domain and range in interval notation.

Assumptions1. The function $h$ is one-to-one, which means that it has an inverse. . The function $h$ is defined as $h(x)=\frac{7x}{x-4}$ .

To find the inverse of a function, we first replace the function notation $h(x)$ with $y$ .
$y=\frac{7x}{x-4}$

Next, we switch the roles of $x$ and $y$ . This means we replace every $x$ in our equation with $y$ and every $y$ with $x$ .
$x=\frac{7y}{y-}$

Now, we solve this equation for $y$ to find $h^{-1}(x)$ . First, multiply both sides by $y-4$ to get rid of the denominator on the right side.
$(y-4)x=7y$

istribute $x$ on the left side.
$yx-4x=7y$

Rearrange the equation to isolate terms with $y$ on one side.
$yx-y=4x$

Factor out $y$ on the left side.
$y(x-7)=4x$

Finally, divide both sides by $x-7$ to solve for $y$ .
$y=\frac{4x}{x-7}$ So, $h^{-1}(x)=\frac{4x}{x-7}$ .

olving the equation gives $x=7$ . This is the value that $x$ cannot take, so the domain of $h^{-}$ is all real numbers except $7$ . In interval notation, this is $(-\infty,7) \cup (7,\infty)$ .

The range of a function is the set of all possible output values (y-values). Since the function $h^{-}$ is a rational function and there is no restriction on the numerator, the range is all real numbers except for the value that makes the numerator zero.
$4x=0$