what is the inverse of (x-3)^2+2
((x-3)squared plus 2))
Do you have a function called
y = (x-3)^2 + 2((x-3)^2 + 2)) ?
then
y = (x-3)^2 + 2(x-3)^2 + 4
= 3(x-3)^2 + 4
to take the inverse,
1. exchange the x and y variables
2. solve the new equation for y
so
x = 3(y-3)^2 + 4
3(y-3)^2 = x-4
(y-3)^2 = (x-4)/3
y-3 = ±√(x-4)/√3
y = 3 ± √(x-4)/√3
It looks to me like your wrote the same function twice.
If you meant to write just
y = (x-3)^2 +2, then:
sqrt(y-2) = x-3
x = sqrt(y-2) + 3
and the inverse function is
y = sqrt(x-2) + 3
To find the inverse of a function, you need to swap the roles of the dependent variable and the independent variable and solve for the original independent variable.
Let's start by rewriting the given function:
y = (x-3)^2 + 2
Step 1: Swap the x and y variables:
x = (y-3)^2 + 2
Step 2: Solve the equation for y.
To do this, we'll begin by isolating the term with the exponent:
x - 2 = (y-3)^2
Next, we can take the square root of both sides:
√(x - 2) = y - 3
Step 3: Solve for y:
y = √(x - 2) + 3
So, the inverse of the function (x-3)^2 + 2 is √(x - 2) + 3.