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Chapter 1: Problem 33

A function \(f: S \rightarrow T\) is specified. Determine if \(f\) is invertible. If it is, state the formula for \(f^{-1}(t) .\) Otherwise, state whether \(f\) fails to be one-to-one, onto, or both. \(S=(4, \infty), T=(1,16 / 15), f(s)=s^{2} /\left(s^{2}-1\right)\)

Short Answer

Expert verified
The function is invertible; the inverse is \(f^{-1}(t) = \sqrt{\frac{t}{1-t}}\).

Step by step solution

01

Check if the function is one-to-one

A function is one-to-one (injective) if each element in the domain maps to a unique element in the codomain. We need to check if \(f(s_1) = f(s_2)\) implies \(s_1 = s_2\) for \(s_1, s_2 \, \in S\.\) Substitute for \(f(s)\):\[\frac{s_1^2}{s_1^2-1} = \frac{s_2^2}{s_2^2-1}\.\] Cross-multiply to obtain \((s_1^2)(s_2^2 - 1) = (s_2^2)(s_1^2 - 1)\.\) Simplify and rearrange to check if \(s_1 = s_2\.\)
02

Simplify and verify one-to-one condition

Expanding the cross-multiplication gives \(s_1^2s_2^2 - s_1^2 = s_2^2s_1^2 - s_2^2\.\) Cancel out \(s_1^2s_2^2\) from both sides to yield \(-s_1^2 = -s_2^2\), which confirms that \(s_1^2 = s_2^2\.\) Since both \(s_1, s_2 > 4\), this implies \(s_1 = s_2\), proving that the function is one-to-one.
03

Check if the function is onto

A function is onto (surjective) if for every element \(t\) in \(T\), there is an element \(s\) in \(S\) such that \(f(s) = t\.\) Set \(t = \frac{s^2}{s^2-1}\) and solve for \(s^2\) in terms of \(t\): \(s^2 = \frac{t}{1-t}\.\) Check the range of \(f(s)\.\)
04

Solve for \(s^2\) and evaluate the range of \(t\)

By solving \(s^2 = \frac{t}{1-t}\) and knowing \(1 < t < \frac{16}{15}\), \(s^2\) must be greater than 4 (since \(s > 4\)). The manipulation shows \(t\) covers the interval \((1, \frac{16}{15})\), thus \(f(s)\) is indeed onto \(T\).
05

Conclusion

Since \(f(s)\) is both one-to-one and onto, it is invertible. To find \(f^{-1}(t)\), use \(s^2 = \frac{t}{1-t}\) and solve for \(s\): \(s = \sqrt{\frac{t}{1-t}}\). Since \(s > 4\), \(f^{-1}(t) = \sqrt{\frac{t}{1-t}}\) is a valid inverse.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

One-to-One Function
When a function is termed *one-to-one*, it means each input value from its domain maps to a unique output in the codomain. This is also known as *injectiveness*. To determine if a function is one-to-one, you would check if, whenever \(f(s_1) = f(s_2)\), it logically follows that \(s_1 = s_2\).For this exercise, the one-to-one property was checked by comparing outputs:
  • We started with the expression \(\frac{s_1^2}{s_1^2-1} = \frac{s_2^2}{s_2^2-1}\).
  • Cross-multiplying yields \(s_1^2 s_2^2 - s_1^2 = s_2^2 s_1^2 - s_2^2\).
  • With simplification, we find \(-s_1^2 = -s_2^2\), leading to \(s_1^2 = s_2^2\). Given that both \(s_1\) and \(s_2\) are greater than 4, it means \(s_1 = s_2\).
This verification ensures that the function does not map different domains to the same range, making it one-to-one.
Onto Function
An *onto* function, or *surjective* function, means that every element in the codomain has at least one element in the domain that maps to it. For a function to be onto, for every value \(t\) in the range, there must be a value \(s\) in the domain such that \(f(s) = t\).In examining if our function is onto, we derived the equation:
  • From \(t = \frac{s^2}{s^2-1}\), solve for \(s^2\), which leads to \(s^2 = \frac{t}{1-t}\).
  • Observe the interval for \(t\), given as \(1 < t < \frac{16}{15}\). For \(f(s)\) to cover this interval fully, \(s^2\) must always remain greater than 4, which is true given the domain \((4, \infty)\).
The function covers the entire range \((1, \frac{16}{15})\), confirming that it is onto.
Inverse Function
If a function is both one-to-one and onto, it can be inverted; that is, an *inverse function* can be determined. The inverse function \(f^{-1}(t)\) allows us to find \(s\) given any valid \(t\).For our function, given that it is determined to be both one-to-one and onto:
  • We solved the equation \(s^2 = \frac{t}{1-t}\) to derive \(s\).
  • The solution is \(s = \sqrt{\frac{t}{1-t}}\).
  • As \(s\) must be greater than 4 (per the domain constraint), this confirms that \(f^{-1}(t) = \sqrt{\frac{t}{1-t}}\) is a valid inverse for \(t\) in \((1, \frac{16}{15})\).
Thus, the invertibility is established and \(f^{-1}(t)\) can be used to find the original input from the output.

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