Problem 56
Question
If \(f(x)\) is one-to-one and \(f(x)\) is never zero, can anything be said about \(h(x)=1 / f(x) ?\) Is it also one-to-one? Give reasons for your answer.
Step-by-Step Solution
Verified Answer
Yes, \(h(x)=\frac{1}{f(x)}\) is one-to-one if \(f(x)\) is one-to-one and never zero.
1Step 1: Define One-to-One Functions
A function \(f(x)\) is said to be one-to-one if it assigns distinct outputs to distinct inputs. In mathematical terms, \(f(a) = f(b)\) implies \(a = b\).
2Step 2: Define the Function h(x)
Given \(h(x) = \frac{1}{f(x)}\), we know that \(h(x)\) is defined for all real numbers where \(f(x) eq 0\), which is valid for this scenario since \(f(x)\) is never zero.
3Step 3: Analyze h(x) for One-to-One Property
To check if \(h(x)\) is one-to-one, consider \(h(a) = h(b)\). Then \(\frac{1}{f(a)} = \frac{1}{f(b)}\) implies \(f(a) = f(b)\). Since \(f(x)\) is one-to-one, it follows that \(a = b\). Thus, \(h(x)\) is also one-to-one.
4Step 4: Conclusion
The analysis shows that because \(f(x)\) is a one-to-one function, \(h(x) = \frac{1}{f(x)}\) is also one-to-one. The reciprocal function does not affect the one-to-one property as long as \(f(x) eq 0\).
Key Concepts
Reciprocal FunctionFunction AnalysisProperties of Functions
Reciprocal Function
When we talk about the reciprocal function, we're essentially looking at flipping the values of a function. If you have a function like \(f(x)\), its reciprocal is given by \(h(x) = \frac{1}{f(x)}\). This means for every value of \(f(x)\), its reciprocal will be one divided by that value.
Understanding how the reciprocal function works helps when considering more complex questions, like whether the new function \(h(x)\), derived from a one-to-one function \(f(x)\), is still one-to-one.
- Keep in mind that the reciprocal of a number is only valid when the original number is not zero because division by zero is undefined.
- That's why, in this scenario, our original function \(f(x)\) must never be zero to ensure that \(h(x)\) is properly defined everywhere.
Understanding how the reciprocal function works helps when considering more complex questions, like whether the new function \(h(x)\), derived from a one-to-one function \(f(x)\), is still one-to-one.
Function Analysis
Function analysis involves exploring the behavior and characteristics of functions, which include properties like domain, range, and the nature of being one-to-one. For the function \(h(x) = \frac{1}{f(x)}\), our analysis starts by understanding its relationship to the original function \(f(x)\).
This approach shows that the transformation to a reciprocal function keeps the function one-to-one. The process of function analysis helps demonstrate this, by carefully checking these relationships, step by step.
- By maintaining the one-to-one nature in \(f(x)\), where different inputs produce distinct outputs, the same property can be examined for \(h(x)\).
- If \(h(a) = h(b)\) implies \(f(a) = f(b)\), then given that \(f(x)\) is one-to-one, we can conclude that \(a = b\).
This approach shows that the transformation to a reciprocal function keeps the function one-to-one. The process of function analysis helps demonstrate this, by carefully checking these relationships, step by step.
Properties of Functions
Properties of functions are crucial because they help us understand how functions behave and relate. Some essential properties include whether a function is one-to-one, continuous, or bounded. All these contribute to how the function can be used and transformed.
Understanding these properties gives us a comprehensive view of how both original and derived functions work, reinforcing the importance of clear definition and behavior for mathematics and applications.
- A one-to-one function, like \(f(x)\), plays a notable role in defining \(h(x) = \frac{1}{f(x)}\), maintaining its one-to-one nature after transformation.
- This is a crucial property because not all transformations preserve every characteristic of the original function, but the reciprocal function does in this case.
- Furthermore, properties such as continuity and existence where the function is non-zero ensures that \(h(x)\) remains well-defined.
Understanding these properties gives us a comprehensive view of how both original and derived functions work, reinforcing the importance of clear definition and behavior for mathematics and applications.
Other exercises in this chapter
Problem 56
Evaluate the integrals. $$\int_{0}^{\pi / 2} 2 \sinh (\sin \theta) \cos \theta d \theta$$
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Find the derivative of \(y\) with respect to the given independent variable. $$y=3^{-x}$$
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Evaluate the integrals. $$\int \frac{6 d r}{\sqrt{4-(r+1)^{2}}}$$
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Use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$y=\sqrt{\left(x^{2}+1\right)(x-1)^{2}}$$
View solution