find the domain and range of f^-1 where f(x)=1/3x+2
If f(x) = 1/(3x+2)
then domain is x ≠ -2/3
range is y ≠ 0
f^-1(x) = (1-2x)/(3x)
and domain is x ≠ 0
range is y ≠ -2/3
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To find the domain and range of the inverse function f^(-1), we need to first find the inverse function of f(x) = (1/3)x + 2.
Step 1: Replace f(x) with y: y = (1/3)x + 2.
Step 2: Swap x and y: x = (1/3)y + 2.
Step 3: Solve the equation for y to find the inverse function:
x - 2 = (1/3)y,
y = 3(x - 2).
Therefore, the inverse function is f^(-1)(x) = 3(x - 2).
Now that we have the inverse function, we can find its domain and range.
Domain: The domain of f^(-1) is the same as the range of the original function f(x).
Since f(x) = (1/3)x + 2 is a linear function, its domain is all real numbers. Therefore, the range of f^(-1) is also all real numbers.
Range: The range of f^(-1) is the same as the domain of the original function f(x).
Since f(x) = (1/3)x + 2 is a linear function, its range is also all real numbers. Therefore, the domain of f^(-1) is also all real numbers.
In summary, the domain and range of f^(-1) are both all real numbers.
To find the domain and range of the inverse function f^-1, we first need to find the inverse function f^-1(x).
Given the function f(x) = 1/3x + 2, let's find its inverse.
Step 1: Replace f(x) with y:
y = 1/3x + 2
Step 2: Swap x and y:
x = 1/3y + 2
Step 3: Solve for y:
x - 2 = 1/3y
Multiply both sides by 3:
3x - 6 = y
So, the inverse function is f^-1(x) = 3x - 6.
Now, let's find the domain and range of f^-1.
The domain of f^-1 is the range of f(x), which is all real numbers since there are no restrictions on x.
Therefore, the domain of f^-1 is (-∞, ∞).
To find the range of f^-1, we can analyze the slope of the function f(x) = 1/3x + 2. Since the slope is positive, the range of f^-1 will also be all real numbers.
Therefore, the range of f^-1 is (-∞, ∞).