For the function f(x)=(3-4x)^2, find f^-1. Determine whether f^1 is a function.
y = (3-4x)^2
find inverse
x = (3-4y)^2
3 -4y = + or - sqrtx
4 y = 3 + or - sqrt x
y = (3/4) +/- sqrt x
this is not a function because there is more than one value of y for eaxh value of x (vertical line rule)
f^-1 is not a function, since f(x) sends two different elements from the domain to the same element of the range.
f must be 1-to-1 for it to have an inverse. f has two distinct branches:
y = (3-4x)^2
3-4x = ±√y
4x = 3±√y
x = (3±√y)/4
So, there is one branch
f^-1 = (3+√y)/4
and the other (3-√y)/4
where y >= 0
To find the inverse of the function f(x) = (3 - 4x)^2, which is denoted as f^(-1), we need to swap the roles of x and y and solve for y.
Step 1: Replace f(x) with y.
y = (3 - 4x)^2
Step 2: Swap x and y.
x = (3 - 4y)^2
Step 3: Solve for y.
Taking the square root of both sides, we get:
√(x) = 3 - 4y
Rearranging the equation to solve for y:
√(x) - 3 = -4y
y = (3 - √(x))/4
Therefore, the inverse function f^(-1)(x) is given by:
f^(-1)(x) = (3 - √(x))/4
To determine whether f^(-1) is a function, we need to check if each input in the domain of f(x) maps to exactly one output in the range of f(x), and vice versa for f^(-1)(x).
In this case, f(x)=(3-4x)^2 is a function since for every x there is exactly one value of y. However, the inverse function f^(-1)(x) = (3 - √(x))/4 is also a function since for every x there is still exactly one value of y. Therefore, both f(x) and f^(-1)(x) are functions.