Find inverse of [1/(x-2)]+3
and
2(x-4)^2 + 5
To find the inverse of a function, you need to swap the x and y values and solve for y. Let's start with the first function:
1. Start with the original function: y = 1/(x-2) + 3.
2. Swap the x and y values: x = 1/(y-2) + 3.
3. Solve for y. To do this, subtract 3 from both sides: x - 3 = 1/(y-2).
4. To get y by itself, take the reciprocal of both sides: 1/(x - 3) = y - 2.
5. Add 2 to both sides to isolate y: 1/(x - 3) + 2 = y.
The inverse of the function [1/(x-2)] + 3 is y = 1/(x - 3) + 2.
Now, let's move on to the second function:
1. Start with the original function: y = 2(x-4)^2 + 5.
2. Swap the x and y values: x = 2(y-4)^2 + 5.
3. Solve for y. Start by subtracting 5 from both sides: x - 5 = 2(y-4)^2.
4. Divide both sides by 2: (x - 5)/2 = (y-4)^2.
5. Take the square root of both sides: √((x - 5)/2) = y - 4.
6. Add 4 to both sides to isolate y: √((x - 5)/2) + 4 = y.
The inverse of the function 2(x-4)^2 + 5 is y = √((x - 5)/2) + 4.
It's important to note that for the functions to have an inverse, they must pass the horizontal line test, meaning that no horizontal line intersects the graph of the function in more than one point.