the inverse of f(x) = 1+ 1/x^2 does not exist, can you make the function so that it's inversed?

The function \( f(x) = 1 + \frac{1}{x^2} \) does not have an inverse because it is not one-to-one (i.e., different values of \( x \) can lead to the same output). Specifically, this function is symmetric around \( x = 0 \), meaning both positive and negative values of \( x \) give the same result, which violates the requirement for a function to have an inverse.

### To modify the function so that it has an inverse:
You can restrict the domain to ensure it is one-to-one. For example, if you limit the domain to \( x > 0 \), then the function becomes one-to-one and can have an inverse.

Thus, if we define the function as \( f(x) = 1 + \frac{1}{x^2} \) but restrict \( x \) to be positive, the function becomes invertible. This restricted function will now be:

\[ f(x) = 1 + \frac{1}{x^2}, \quad \text{for } x > 0. \]

Now, we can find the inverse of this restricted function.

### Finding the inverse:

1. Start with \( y = 1 + \frac{1}{x^2} \).
2. Subtract 1 from both sides: \( y - 1 = \frac{1}{x^2} \).
3. Take the reciprocal of both sides: \( \frac{1}{y-1} = x^2 \).
4. Take the square root of both sides: \( x = \frac{1}{\sqrt{y-1}} \), since \( x > 0 \).

Thus, the inverse function is:

\[ f^{-1}(y) = \frac{1}{\sqrt{y-1}}, \quad \text{for } y > 1. \]

This is the inverse of the original function, with the restriction \( x > 0 \) on the domain of the original function, and \( y > 1 \) for the range of the inverse function.

Would you like to explore more details or have any questions on this? Here are five related questions to think about:

1. What happens to the graph of the function when we restrict the domain to \( x > 0 \)?
2. Can you restrict the domain to \( x < 0 \) and still find an inverse?
3. How would the inverse function change if the function had a different form, such as \( f(x) = 2 + \frac{1}{x^2} \)?
4. What does it mean for a function to be one-to-one, and why is it necessary for inverses?
5. How would you graph the inverse function?

**Tip**: When finding an inverse, always check whether the function is one-to-one by examining if different inputs give unique outputs.

The inverse of f(x) = 1 + 1/x^2 does not exist, can you make the function so that it's inversed?

Learn how to make the function f(x) = 1 + 1/x^2 invertible by restricting its domain to x > 0. Explore the detailed step-by-step solution including key concepts like one-to-one functions, domain restrictions, and inverse function calculations. Suitable for high school students in Grades 10-12.

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