Find inverse of [1/(x-2)]+3
and
2(x-4)^2 + 5
To find the inverse of a function, we need to interchange the x and y variables and solve for y.
1. Inverse of [1/(x-2)]+3:
Step 1: Replace f(x) with y:
y = [1/(x-2)] + 3
Step 2: Swap x and y:
x = [1/(y-2)] + 3
Step 3: Solve for y:
x - 3 = 1/(y-2)
Cross-multiplying:
x - 3 = y - 2
Step 4: Isolate y:
y = x - 1
Therefore, the inverse of [1/(x-2)] + 3 is y = x - 1.
2. Inverse of 2(x-4)^2 + 5:
Step 1: Replace f(x) with y:
y = 2(x-4)^2 + 5
Step 2: Swap x and y:
x = 2(y-4)^2 + 5
Step 3: Solve for y:
x - 5 = 2(y-4)^2
Divide both sides by 2:
(x - 5) / 2 = (y-4)^2
Step 4: Take the square root of both sides:
±√[(x - 5) / 2] = y - 4
Step 5: Solve for y:
y = ±√[(x - 5) / 2] + 4
Therefore, the inverse of 2(x-4)^2 + 5 is y = ±√[(x - 5) / 2] + 4.