1. Find f^-1 for the function f(x)=3srtx-2 +8
2. Find the domain and range of f^-1 where f(x) = 1/3x+2
1. swap variables and solve for the new y = f^-1(x)
x = 3√(y-2)+8
x-8 = 3√(y-2)
finish it up. Check your answer using the fact that if g(x) = f^-1(x) then
f(g(x)) = g(f(x)) = x
#2 well, f^-1(x) = (1/x - 2)/3
so, what do you think? Start by picturing (or, heck, actually graphing!)
y = 1/x
and then shift it and scale it.
To find the inverse of a function, you need to interchange x and y and solve for y. Let's go through each question step by step.
1. Find f^-1 for the function f(x) = 3√(x-2) + 8:
Step 1: Replace f(x) with y: y = 3√(x-2) + 8.
Step 2: Interchange x and y: x = 3√(y-2) + 8.
Step 3: Solve for y: Subtract 8 from both sides and isolate the square root term:
x - 8 = 3√(y-2).
Step 4: Isolate the square root term: Divide by 3:
(x - 8) / 3 = √(y-2).
Step 5: Square both sides to eliminate the square root:
[((x - 8) / 3)]^2 = y - 2.
Step 6: Simplify and solve for y:
[((x - 8) / 3)]^2 + 2 = y.
So, the inverse function f^-1(x) for f(x) = 3√(x-2) + 8 is:
f^-1(x) = [((x - 8) / 3)]^2 + 2.
2. Find the domain and range of f^-1 where f(x) = (1/3)x + 2:
To find the domain and range, we need to examine the restrictions and potential values for x and y.
Domain: The domain refers to all the possible values that x can take in the function. For f^-1 to be defined, the original function f(x) must be one-to-one, meaning that each x-value has a unique corresponding y-value. In the case of f(x) = (1/3)x + 2, it is a linear function, and every x-value has a unique y-value. Thus, the domain of f^-1 is all real numbers.
Range: The range represents all the potential y-values that f^-1 can take. Since the original function f(x) = (1/3)x + 2 is an increasing linear function, it means that the inverse function f^-1 will also be increasing. Therefore, the range of f^-1 is all real numbers.
In summary:
1. The inverse function f^-1(x) for f(x) = 3√(x-2) + 8 is f^-1(x) = [((x - 8) / 3)]^2 + 2.
2. The domain of f^-1 is all real numbers, and the range of f^-1 is also all real numbers.