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For the function, find the inverse and the domain and range of the function and its inverse. Determine whether the inverse is a function. f(x)=\sqrt(x+1)

User Ilshidur
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To start this, let's identify a crucial point about the function f(x) = √(x+1):

- This function will not accept negative values for x because it is not feasible to compute the square root of a negative number. So, the domain for f(x) = √(x+1) will be x >= -1. The function f(x) will only return values equal to or greater than zero, so the range will be y >= 0.

From there, we will start the process to find the inverse of the function:

1. First, swap the output (y) with the input (x). So it becomes: y = √(x+1)

2. Then, square both sides of the equation to get rid of the square root operation: y² = x + 1

3. Solve the equation for x to get: x = y² - 1

4. That tells us the inverse function is f⁻¹(y) = y² - 1

Now, we can also derive the domain and range of the inverse function by mirroring that of the original function:

- The domain of f⁻¹, the inverse function, will be y >= 0, all the values greater than or equal to 0.

- The range will be x >= -1, all the values greater than or equal to -1.

The inverse function does pass the vertical line test. That is, for every input y in the domain, there's exactly one output x in the range. So the inverse is indeed a function.

In summary, The inverse function of f(x) = √(x+1) is f⁻¹(y) = y² - 1. The domain of the original function is x >= -1 and the range is y >= 0. The domain of the inverse function is y >= 0 and the range is x >= -1. The inverse is a function.

User HMHero
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