Solved on Jan 25, 2024

Find the inverse function of $y=\sqrt{x-1}+3$

STEP 1

Assumptions
1. The given relation is $y = \sqrt{x - 1} + 3$ .
2. To find the inverse, we need to solve for $x$ in terms of $y$ .
3. The domain of the function $y = \sqrt{x - 1} + 3$ is $x \geq 1$ since the square root function is defined for non-negative numbers.

STEP 2

To find the inverse of the relation, we first express $y$ in terms of $x$ and then swap $x$ and $y$ .
$y = \sqrt{x - 1} + 3$

STEP 3

Subtract 3 from both sides of the equation to isolate the square root term.
$y - 3 = \sqrt{x - 1}$

STEP 4

Square both sides of the equation to eliminate the square root. Remember that squaring both sides of an equation is a valid operation, but we must be cautious of any restrictions that may arise from the domain of the original function.
$(y - 3)^2 = (\sqrt{x - 1})^2$

STEP 5

Simplify the right side of the equation by removing the square root.
$(y - 3)^2 = x - 1$

STEP 6

Add 1 to both sides of the equation to solve for $x$ .
$(y - 3)^2 + 1 = x$

STEP 7

Now, we have $x$ expressed in terms of $y$ . To write the inverse relation, we swap $x$ and $y$ .
$y = (x - 3)^2 + 1$

STEP 8

The inverse relation of $y = \sqrt{x - 1} + 3$ is $y = (x - 3)^2 + 1$ , where the domain is $x \geq 3$ to match the range of the original function and ensure the inverse is a function.

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Find the inverse function of $y=\sqrt{x-1}+3$

STEP 1

Assumptions
1. The given relation is $y = \sqrt{x - 1} + 3$ .
2. To find the inverse, we need to solve for $x$ in terms of $y$ .
3. The domain of the function $y = \sqrt{x - 1} + 3$ is $x \geq 1$ since the square root function is defined for non-negative numbers.

STEP 2

To find the inverse of the relation, we first express $y$ in terms of $x$ and then swap $x$ and $y$ .
$y = \sqrt{x - 1} + 3$

STEP 3

Subtract 3 from both sides of the equation to isolate the square root term.
$y - 3 = \sqrt{x - 1}$

STEP 4

Square both sides of the equation to eliminate the square root. Remember that squaring both sides of an equation is a valid operation, but we must be cautious of any restrictions that may arise from the domain of the original function.
$(y - 3)^2 = (\sqrt{x - 1})^2$

STEP 5

Simplify the right side of the equation by removing the square root.
$(y - 3)^2 = x - 1$

STEP 6

Add 1 to both sides of the equation to solve for $x$ .
$(y - 3)^2 + 1 = x$

STEP 7

Now, we have $x$ expressed in terms of $y$ . To write the inverse relation, we swap $x$ and $y$ .
$y = (x - 3)^2 + 1$

STEP 8

The inverse relation of $y = \sqrt{x - 1} + 3$ is $y = (x - 3)^2 + 1$ , where the domain is $x \geq 3$ to match the range of the original function and ensure the inverse is a function.

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Find the inverse function of y=x−1+3y=\sqrt{x-1}+3y=x−1​+3

STEP 1

STEP 2

To find the inverse of the relation, we first express y y y in terms of x x x and then swap x x x and y y y.y=x−1+3 y = \sqrt{x - 1} + 3 y=x−1​+3

STEP 3

Subtract 3 from both sides of the equation to isolate the square root term.y−3=x−1 y - 3 = \sqrt{x - 1} y−3=x−1​

STEP 4

STEP 5

Simplify the right side of the equation by removing the square root.(y−3)2=x−1 (y - 3)^2 = x - 1 (y−3)2=x−1

STEP 6

Add 1 to both sides of the equation to solve for x x x.(y−3)2+1=x (y - 3)^2 + 1 = x (y−3)2+1=x

STEP 7

Now, we have x x x expressed in terms of y y y. To write the inverse relation, we swap x x x and y y y.y=(x−3)2+1 y = (x - 3)^2 + 1 y=(x−3)2+1

STEP 8

The inverse relation of y=x−1+3 y = \sqrt{x - 1} + 3 y=x−1​+3 is y=(x−3)2+1 y = (x - 3)^2 + 1 y=(x−3)2+1, where the domain is x≥3 x \geq 3 x≥3 to match the range of the original function and ensure the inverse is a function.

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Find the inverse function of $y=\sqrt{x-1}+3$

To find the inverse of the relation, we first express $y$ in terms of $x$ and then swap $x$ and $y$ .
$y = \sqrt{x - 1} + 3$

Subtract 3 from both sides of the equation to isolate the square root term.
$y - 3 = \sqrt{x - 1}$

Simplify the right side of the equation by removing the square root.
$(y - 3)^2 = x - 1$

Add 1 to both sides of the equation to solve for $x$ .
$(y - 3)^2 + 1 = x$

Now, we have $x$ expressed in terms of $y$ . To write the inverse relation, we swap $x$ and $y$ .
$y = (x - 3)^2 + 1$

The inverse relation of $y = \sqrt{x - 1} + 3$ is $y = (x - 3)^2 + 1$ , where the domain is $x \geq 3$ to match the range of the original function and ensure the inverse is a function.