Problem 57
Question
In Exercises 57-60, determine all values of \(x\) at which the function is discontinuous. \(f(x)=\frac{2 x}{x^{2}-1}\)
Step-by-Step Solution
Verified Answer
The function \(f(x) = \frac{2x}{x^2 - 1}\) is discontinuous at x = 1 and x = -1.
1Step 1: Identify the Denominator
In our given function, f(x) = \(\frac{2x}{x^2 - 1}\), the denominator is (x^2 - 1).
2Step 2: Set the Denominator Equal to Zero
To find where the function might be discontinuous, we need to solve the equation x^2 - 1 = 0.
3Step 3: Solve the Equation
We're solving the equation: x^2 - 1 = 0
Using the property of difference of squares, we can factor the equation like this:
(x - 1)(x + 1) = 0
Now that we've factored, we solve the equation by setting each of the factors equal to 0:
x - 1 = 0 => x = 1
x + 1 = 0 => x = -1
So, the potential points of discontinuity are x = 1 and x = -1.
4Step 4: Verify the Discontinuities
We don't need to verify the discontinuities, because any value that makes the denominator equal to 0 results in an undefined fraction, and therefore the function will be discontinuous at x = 1 and x = -1.
In conclusion, the function f(x) = \(\frac{2x}{x^2 - 1}\) is discontinuous for x = 1 and x = -1.
Key Concepts
Difference of SquaresRational FunctionsUndefined Expressions
Difference of Squares
The difference of squares is a fundamental algebraic concept used to simplify various expressions. It refers to any expression in the form \(a^2 - b^2\). This expression can be factored as \((a - b)(a + b)\), which is particularly useful in solving equations.
In the context of our function, \(f(x) = \frac{2x}{x^2 - 1}\), we encounter a difference of squares in the denominator: \(x^2 - 1\). Using the difference of squares formula, we can factor it into \((x - 1)(x + 1)\).
Understanding how to factor difference of squares helps in identifying when and where potential discontinuities occur. By applying this knowledge, we set each factor equal to zero: \(x - 1 = 0\) and \(x + 1 = 0\), finding that \(x = 1\) and \(x = -1\) are where the function could be discontinuous.
In the context of our function, \(f(x) = \frac{2x}{x^2 - 1}\), we encounter a difference of squares in the denominator: \(x^2 - 1\). Using the difference of squares formula, we can factor it into \((x - 1)(x + 1)\).
Understanding how to factor difference of squares helps in identifying when and where potential discontinuities occur. By applying this knowledge, we set each factor equal to zero: \(x - 1 = 0\) and \(x + 1 = 0\), finding that \(x = 1\) and \(x = -1\) are where the function could be discontinuous.
Rational Functions
Rational functions are fractions where both the numerator and the denominator are polynomials. They often present unique challenges when it comes to finding discontinuities. The function in our example, \(f(x) = \frac{2x}{x^2 - 1}\), is a typical rational function.
To understand where these functions might be discontinuous, observe the denominator: it's crucial because when the denominator equals zero, the function becomes undefined. In this case, you will set \(x^2 - 1 = 0\) to identify problem points. After factoring it as \((x - 1)(x + 1)\), we find the discontinuities at \(x = 1\) and \(x = -1\).
Analyzing rational functions enables us to understand where undefined expressions might occur, leading to discontinuities. By meticulously solving these points, you can unlock the entire behavior of the function.
To understand where these functions might be discontinuous, observe the denominator: it's crucial because when the denominator equals zero, the function becomes undefined. In this case, you will set \(x^2 - 1 = 0\) to identify problem points. After factoring it as \((x - 1)(x + 1)\), we find the discontinuities at \(x = 1\) and \(x = -1\).
Analyzing rational functions enables us to understand where undefined expressions might occur, leading to discontinuities. By meticulously solving these points, you can unlock the entire behavior of the function.
Undefined Expressions
Undefined expressions are key indicators of where a function isn't continuous. For rational functions, these occur when the denominator is zero, leading the fraction to be undefined. Identifying places where expressions are undefined helps in determining discontinuity.
In our function, \(f(x) = \frac{2x}{x^2 - 1}\), the denominator \(x^2 - 1\) zeroes out at \(x = 1\) and \(x = -1\), causing the function to be undefined. This is because no real number can divide zero, making the whole expression undefined at these points.
Understanding undefined expressions involves setting the denominator equal to zero and solving for \(x\). It's a crucial step to localize discontinuities and understand the function's behavior fully. Remember, whenever the denominator hits zero, you'll find a point of discontinuity due to the undefined nature of the expressions.
In our function, \(f(x) = \frac{2x}{x^2 - 1}\), the denominator \(x^2 - 1\) zeroes out at \(x = 1\) and \(x = -1\), causing the function to be undefined. This is because no real number can divide zero, making the whole expression undefined at these points.
Understanding undefined expressions involves setting the denominator equal to zero and solving for \(x\). It's a crucial step to localize discontinuities and understand the function's behavior fully. Remember, whenever the denominator hits zero, you'll find a point of discontinuity due to the undefined nature of the expressions.
Other exercises in this chapter
Problem 57
Find the point(s) on the graph of the function \(f(x)=\) \(\left(x^{2}+6\right)(x-5)\) where the slope of the tangent line is equal to \(-2\).
View solution Problem 57
Sketch the graph of the function \(f(x)=|x+1|\) and show that the function does not have a derivative at \(x=-1\).
View solution Problem 57
Find the indicated limit, if it exists. \(\lim _{x \rightarrow-2} \frac{x^{2}-x-6}{x^{2}+x-2}\)
View solution Problem 58
Find the point \((s)\) on the graph of the function \(f(x)=\frac{x+1}{x-1}\) where the slope of the tangent line is equal to \(-\frac{1}{2}\).
View solution