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Chapter 0: Problem 70

\(f(x)=\frac{x-5}{\sqrt{x^{2}-9}}\)

Short Answer

Expert verified
The simplified version of the function is \(f(x)= \frac{x-5}{\sqrt{(x-3)(x+3)}}\). The domain of the function is \((-鈭�, -3) U (3, 鈭�)\). Graphic representation of this function will have vertical asymptotes at \(x = 3\) and \(x = -3\).

Step by step solution

01

Simplifying the denominator

The denominator is a square root of \(x^{2}-9\), which is a difference of squares, so it could be rewritten as \(\sqrt{(x-3)(x+3)}\). This might simplify further calculations.
02

Finding the domain

The denominator cannot be zero or a negative under the square root as it would make the function undefined. This leads to the following conclusions: 1. \(x^{2}-9 = 0\) gives us \(x = 3\) and \(x = -3\). Thus, \(x \neq 3\) and \(x \neq -3\). 2. \(x^{2}-9 > 0\) gives us \(x > 3\) or \(x < -3\). Therefore, the domain of this function is \((-鈭�, -3) U (3, 鈭�)\).
03

Plotting the function

To understand the behavior of the function, it would be advisable to plot it using the domain obtained in step 2. Due to the square root in the denominator and fraction form of the function, the graph would have vertical asymptotes at \(x = 3\) and \(x = -3\). For \(x < -3\) or \(x > 3\), the function will yield real values which can be plotted on the graph.

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Key Concepts

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

Domain of a Function
Understanding the domain of a function is crucial in mathematics. The domain is the set of all possible input values (usually denoted as 'x') for which the function is defined.
For the function given, \[f(x)=\frac{x-5}{\sqrt{x^{2}-9}}\]we need to ensure that the denominator is never zero or negative.
This means the expression inside the square root \[x^2 - 9\]must be greater than zero.
Solving \[x^2 - 9 = 0\] gives us the points \[x = 3\] and \[x = -3\].
This provides restrictions, so \[x eq 3\] and \[x eq -3\].
  • Furthermore, \[x^2 - 9 > 0\] implies \[x > 3\] or \[x < -3\]. Therefore, the function is defined for input values of \[(-鈭�, -3) U (3, 鈭�)\],which forms the domain of the given function.
Difference of Squares
The difference of squares is a powerful tool in algebra. It applies to expressions like \[x^2 - a^2\] which can be factored easily into \[(x-a)(x+a)\].
For the function \[f(x)=\frac{x-5}{\sqrt{x^{2}-9}}\], the expression in the denominator \[x^2 - 9\] is a difference of squares and can be rewritten as \[(x-3)(x+3)\].
This factorization helps in simplifying and understanding the behavior of the function.
  • By realizing this pattern, it becomes clear why the expression becomes zero at \[x = 3\] and \[x = -3\].
  • It's essential to factor differences of squares whenever possible, especially when simplifying expressions or solving equations.
Vertical Asymptotes
Vertical asymptotes occur in a graph of a function where the value tends to infinity (or negative infinity) as it approaches a particular point. These are often points where the function is undefined.
In our example, the function \[f(x)=\frac{x-5}{\sqrt{x^{2}-9}}\],the vertical asymptotes occur at \[x = 3\] and \[x = -3\].
This is because the denominator equals zero at these values, making the function undefined.
  • To visualize it, as \[x\] approaches \[3\] from the left or right, the values of \[f(x)\] will sharply tend towards infinity.
  • Similarly, the behavior is mirrored at \[x = -3\].
This creates a break or gap in the graph at these points, indicating the presence of vertical asymptotes.
Graphing helps depict how the function behaves in regions around the asymptotes, giving a fuller picture of the function's behavior.

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Most popular questions from this chapter

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