Find the inverse of the function f(x) = 2/3x + 5. Mention if there is any restriction on domain of f^ -1
y = 2/3 x + 5
y-5 = 2/3 x
x = 3/2 (y-5)
so, f^-1(x) = 3/2 (x-5)
as with all polynomials, the domain is all real numbers.
To find the inverse of the function f(x) = 2/3x + 5, we can follow these steps:
Step 1: Replace f(x) with y:
y = 2/3x + 5
Step 2: Swap x and y:
x = 2/3y + 5
Step 3: Solve for y:
x - 5 = 2/3y
Multiply both sides by 3/2 to isolate y:
(3/2)(x - 5) = y
Simplify:
3/2x - 15/2 = y
So, the inverse function is:
f^ -1(x) = 3/2x - 15/2
Now, let's address any restrictions on the domain of f^ -1.
The restriction on the domain of f^ -1 is related to the original function f(x). In this case, there is no restriction on the domain of f(x), which means that the inverse function f^ -1(x) is defined for all real values of x. In other words, there are no values of x for which the original function f(x) is not invertible. Therefore, the domain of f^ -1 is (-∞, +∞), or all real numbers.