Problem 29 - Thomas Calculus in SI Units

Question

Gives a formula for a function \(y=f(x) .\) In each case, find \(f^{-1}(x)\) and identify the domain and range of \(f^{-1} .\) As a check, show that \(f\left(f^{-1}(x)\right)=f^{-1}(f(x))=x\). $$f(x)=1 / x^{2}, \quad x>0$$

Step-by-Step Solution

Answer

The inverse is \( f^{-1}(x) = \frac{1}{\sqrt{x}} \), with domain \( x > 0 \) and range \( y > 0 \).

Step 1: Understand the Original Function

The given function is \( f(x) = \frac{1}{x^2} \) with the restriction \( x > 0 \). This means that \( f(x) \) takes positive real numbers as input and outputs real numbers that are greater than zero.

Step 2: Rearrange the Function for the Inverse

To find \( f^{-1}(x) \), start by expressing the function \( y = \frac{1}{x^2} \) in terms of \( x \). This needs to be rearranged to solve for \( x \) in terms of \( y \).

Step 3: Swap Variables

From \( y = \frac{1}{x^2} \), swap \( x \) and \( y \) to start finding the inverse: \( x = \frac{1}{y^2} \). Solve this equation for \( y \) to find \( y = f^{-1}(x) \).

Step 4: Solve for the Inverse Function

Taking reciprocals on both sides gives \( y = \frac{1}{\sqrt{x}} \). Thus, \( f^{-1}(x) = \frac{1}{\sqrt{x}} \). Since \( x > 0 \) in \( f(x) \), \( f^{-1}(x) \) must also take positive values for the domain.

Step 5: Determine the Domain and Range of f^{-1}

The domain of \( f^{-1}(x) \) is \( x > 0 \), since \( \sqrt{x} \) is only defined for positive \( x \). The range of \( f^{-1}(x) \) is also \( y > 0 \), as it outputs positive numbers.

Step 6: Verify f(f^{-1}(x))=x

Substitute \( f^{-1}(x) \) into \( f(x) \): \( f\left(\frac{1}{\sqrt{x}}\right) = \frac{1}{\left(\frac{1}{\sqrt{x}}\right)^2} = x\).

Step 7: Verify f^{-1}(f(x))=x

Substitute \( f(x) \) into \( f^{-1}(x) \): \( f^{-1}\left(\frac{1}{x^2}\right) = \frac{1}{\sqrt{\frac{1}{x^2}}} = x \).

Step 8: Conclusion

The inverse function \( f^{-1}(x) = \frac{1}{\sqrt{x}} \) and it is verified that \( f(f^{-1}(x))=x \) and \( f^{-1}(f(x))=x \). The domain and range of \( f^{-1} \) are both \( x > 0 \).

Key Concepts

Domain and Range

When dealing with inverse functions, it's important to understand the concepts of domain and range as they lay the foundation for how functions behave and interact. The domain refers to the set of all possible input values for the function. For example, in our original function

• The domain of f(x) is \( x > 0 \), meaning it only accepts positive real numbers.

The range, on the other hand, is the set of all possible output values. It's what the function can produce, or map to, from its inputs.

• For our function \( f(x) = \frac{1}{x^2} \), since we only input positive x, it gives outputs greater than zero. Thus, its range is also \( y > 0 \).

When we consider the inverse function \( f^{-1}(x) = \frac{1}{\sqrt{x}} \), we switch roles:

• The domain of the inverse function is the range of the original function, \( x > 0 \).
• Similarly, the range of the inverse function becomes \( y > 0 \), as it maps positive outputs to inputs.

Function Composition

Function composition involves combining two functions such that the output of one function becomes the input of the other. This process allows us to explore the relationships between a function and its inverse. In our example,

• We use function composition to show that performing a function and then its inverse returns us to our starting value: \( f(f^{-1}(x)) = x \).
• This can be understood by looking at the operations: taking the inverse and then the original cancels out any transformation.

The same occurs in the reverse: applying the inverse after the original should also revert the output back to the initial input, stated as \( f^{-1}(f(x)) = x \).

• This is because if you input \( x \) into the original function and then input the result into its inverse, you should logically end up with your initial input value, ensuring the properties of the inverse function.

This mathematical exploration form a crucial verification step, confirming the relationship between \( f \) and \( f^{-1} \).

Mathematical Verification

Verification is an essential step in confirming that we have correctly deduced the inverse. It involves checking through calculation to ensure the functions truly behave as expected. For verification:

• We first substitute \( f^{-1}(x) \) back into \( f(x) \) and simplify. This should give us \( x \), the original input: \[ f\left(\frac{1}{\sqrt{x}}\right) = \frac{1}{\left(\frac{1}{\sqrt{x}}\right)^2} = x. \]
• Then, substitute \( f(x) \) into \( f^{-1}(x) \) yielding \( x \) again: \[ f^{-1}\left(\frac{1}{x^2}\right) = \frac{1}{\sqrt{\frac{1}{x^2}}} = x. \]

These checks, known as the inverse properties, confirm that \( f^{-1} \) is correctly derived. If these relationships hold true, we can be confident in our solution.

• This ensures that the reverse process truly undoes the original functions, and validates the calculated inverse.

Through mathematical verification, we verify both the theoretical understanding and practical application of inverse functions in mathematics.