Chapter 6: Problem 55
Simplify the given expressions. If \(f(x)=x-\frac{1}{x},\) find \(f(a+1).\)
Short Answer
Expert verified
\(f(a+1) = a + 1 - \frac{1}{a+1}\).
Step by step solution
01
Understand the Function
The function given is \(f(x) = x - \frac{1}{x}\). It takes an input \(x\), subtracts its reciprocal from \(x\) to yield the output.
02
Substitute the Expression into the Function
To find \(f(a+1)\), substitute \(a+1\) for \(x\) in the function \(f(x) = x - \frac{1}{x}\). Thus, you get \(f(a+1) = (a+1) - \frac{1}{a+1}\).
03
Simplify the Expression
Start with the expression \((a+1) - \frac{1}{a+1}\). The expression \(a+1\) is already simplified. The term \(\frac{1}{a+1}\) remains as it is. Therefore, the simplified output is \(a+1 - \frac{1}{a+1}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Functions
In mathematics, a function is a relation between a set of inputs and a set of permissible outputs, with each input being related to exactly one output. Functions are often expressed as equations. For example, the function given in our problem is \( f(x) = x - \frac{1}{x} \).
This means for each input \( x \), the function calculates the output by subtracting the reciprocal of \( x \) from \( x \) itself. Functions are essential because they help us understand and model relationships between variables in a precise way.
Here are a few key points about functions:
This means for each input \( x \), the function calculates the output by subtracting the reciprocal of \( x \) from \( x \) itself. Functions are essential because they help us understand and model relationships between variables in a precise way.
Here are a few key points about functions:
- Domain: The set of all possible input values (\( x \)) for the function.
- Range: The set of all possible output values the function can produce.
- Function Notation: Typically written as \( f(x), g(x), h(x) \), meaning "function of \( x \)."
Substitution
Substitution is a basic yet vital concept in mathematics that makes solving equations more manageable.
It involves replacing a variable or an expression in an equation with another equivalent expression or value to simplify or solve it. In the context of our exercise, we substituted \( x \) with \( a+1 \) in the function \( f(x) = x - \frac{1}{x} \) to find \( f(a+1) \).
Let's dig deeper:
It involves replacing a variable or an expression in an equation with another equivalent expression or value to simplify or solve it. In the context of our exercise, we substituted \( x \) with \( a+1 \) in the function \( f(x) = x - \frac{1}{x} \) to find \( f(a+1) \).
Let's dig deeper:
- When we perform substitution, we replace every instance of a variable with the expression we want to substitute. In our example, everywhere we see \( x \), we replace it with \( a+1 \).
- This substitution process helps us understand how the function behaves with different inputs, and it's especially useful in scenarios where we have specific inputs to evaluate, like in this problem.
- After substitution, we proceed with simplifying the expression to achieve the final output, here resulting in \( a+1 - \frac{1}{a+1} \).
Reciprocal Functions
Reciprocal functions are a fascinating area of functions where we deal with the inverse or opposite of a value.
The reciprocal of a number \( x \) is represented as \( \frac{1}{x} \), meaning dividing 1 by \( x \). Reciprocal functions often invert or flip usual operations, offering unique insights into relationships.In our exercise, \( \frac{1}{x} \) plays a critical role in the function \( f(x) = x - \frac{1}{x} \).
Consider these factors:
The reciprocal of a number \( x \) is represented as \( \frac{1}{x} \), meaning dividing 1 by \( x \). Reciprocal functions often invert or flip usual operations, offering unique insights into relationships.In our exercise, \( \frac{1}{x} \) plays a critical role in the function \( f(x) = x - \frac{1}{x} \).
Consider these factors:
- Reciprocal functions can help us explore opposite transformations of certain mathematical operations, leading to inversions in behavior.
- For \( x eq 0 \), \( x \) and \( \frac{1}{x} \) represent an interesting case where multi-dimensional relationships (like hyperbolas) emerge upon graphing.
- Reciprocal values impact the behavior of equations, and their inclusion often results in intricate function dynamics.