Answer:
1. y = 1/5^(1-x) + 2
The inverse function, denoted as f^-1(x), can be found by swapping the x and y variables and solving for y.
So, x = 1/5^(1-y) + 2
To find the inverse, we need to solve for y:
y = 1 - log(5, x-2)
Note that the inverse only exists if the original function is a one-to-one function, meaning that it assigns a unique output to each input.
The domain of f is all x such that x > 2. And the range of f is all y such that y > 2. The domain of f^-1 is all x such that x > 2 and the range of f^-1 is all y such that y > 1.
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2. 1 - 5^x/5^(x-1)
The inverse function, denoted as f^-1(x), can be found by swapping the x and y variables and solving for y.
So, x = 1 - 5^y/5^(y-1)
To find the inverse, we need to solve for y:
y = log(5, 1/(x+1)) + 1
Note that the inverse only exists if the original function is a one-to-one function, meaning that it assigns a unique output to each input.
The domain of f is all x such that x < 1. And the range of f is all y such that 0 < y < 1. The domain of f^-1 is all x such that x < 1 and the range of f^-1 is all y such that y > 1.
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3. y = log1/5(x+1/2)-1
The inverse function, denoted as f^-1(x), can be found by swapping the x and y variables and solving for y.
So, x = log1/5(y+1) - 1/2
To find the inverse, we need to solve for y:
y = 5^(x+1/2) - 1
Note that the inverse only exists if the original function is a one-to-one function, meaning that it assigns a unique output to each input.
The domain of f is all x such that x exists. And the range of f is all y such that y > -1. The domain of f^-1 is all x such that x exists and the range of f^-1 is all y such that y > 0.
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In summary,
1. f^-1(x) = 1 - log(5, x-2) Domain: x>2, Range: y>2.
2. f^-1(x) = log(5, 1/(x+1)) + 1 Domain: x<1, Range: 0<y<1.
3. f^-1(x) = 5^(x+1/2) - 1 Domain: x exists, Range: y>-1.