Problem 47
Question
Use a graphing utility to graph the function. Use the graph to determine any \(x\) -value(s) at which the function is not continuous. Explain why the function is not continuous at the \(x\) -value(s). \(h(x)=\frac{1}{x^{2}-x-2}\)
Step-by-Step Solution
Verified Answer
The function \(h(x)=\frac{1} {x^{2}-x-2}\) is not continuous at \(x = 2\) and \(x = -1\) as there are vertical asymptotes at these points.
1Step 1: Factorize the denominator
Start by factoring the denominator in the function to simplify the continuity analysis. \(x^{2}-x-2\) can be factored into \((x - 2)(x + 1)\). So, the function becomes \(h(x)=\frac{1} {(x - 2)(x +1)}.\)
2Step 2: Graph the Function
To plot graph of a function, use a graphing utility or software. In this function, look for any gaps, jumps or vertical asymptotes, indicating discontinuity.
3Step 3: Identify Discontinuities
Determining where the function is not defined will provide the \(x\)-values where the function is not continuous. For the function \(h(x)=\frac{1} {(x - 2)(x +1)}\), the function is undefined (unbounded) when the denominator equals zero. Solve \(x - 2 = 0\) and \(x + 1 = 0\) for \(x\), leading to \(x = 2\) and \(x = -1\). Therefore, the function is not continuous at \(x = 2\) and \(x = -1\). The discontinuities are due to vertical asymptotes at these values of \(x\).
4Step 4: Explain the discontinuity
The function has vertical asymptotes at \(x = 2\) and \(x = -1\). These are values where the function is undefined; therefore, a vertical asymptote occurs, resulting in a discontinuity in the graph. In other words, there is a jump in the function values at these points.
Key Concepts
Continuity of FunctionsVertical AsymptotesFactoring PolynomialsGraphing Utilities
Continuity of Functions
Continuity in functions is the property that describes how well-connected the values of a function are as we trace along its graph. When we say a function is continuous at a particular value of \( x \), it means that as \( x \) approaches that value, the function's output approaches a well-defined limit, and there is no sudden jump or gap in the function.
In the context of the exercise, continuity is broken at the points where the denominator of our function \( h(x) \) becomes zero. These points are \( x = 2 \) and \( x = -1 \). When approaching these values, the function's output grows without bound, leading to an infinite discontinuity, which is a type of discontinuity characterized by a function seemingly leaping to infinity.
In the context of the exercise, continuity is broken at the points where the denominator of our function \( h(x) \) becomes zero. These points are \( x = 2 \) and \( x = -1 \). When approaching these values, the function's output grows without bound, leading to an infinite discontinuity, which is a type of discontinuity characterized by a function seemingly leaping to infinity.
Vertical Asymptotes
Vertical asymptotes are lines that a function approaches as the variable approaches a specific value but never actually reaches. These lines represent values for which the function's limit is infinity. They occur in places where the function is undefined and typically result when we have a zero in the denominator of a rational function, but not in the numerator.
For the given exercise function \( h(x) = \frac{1}{x^{2}-x-2} \), the vertical asymptotes are at the values of \( x \) that cause the denominator to be zero. After factoring the polynomial, we find that the function has vertical asymptotes at \( x = 2 \) and \( x = -1 \), since these values make the function take on the form which tends towards infinity.
For the given exercise function \( h(x) = \frac{1}{x^{2}-x-2} \), the vertical asymptotes are at the values of \( x \) that cause the denominator to be zero. After factoring the polynomial, we find that the function has vertical asymptotes at \( x = 2 \) and \( x = -1 \), since these values make the function take on the form which tends towards infinity.
Factoring Polynomials
Factoring polynomials is a critical algebraic process used to simplify expressions and solve equations. It involves breaking down a complex polynomial into a product of simpler polynomials, usually linear or quadratic factors that are easier to manage.
In our exercise, the polynomial \( x^{2}-x-2 \) was factored into \( (x - 2)(x + 1) \), simplifying the identification of discontinuities in the rational function. Each factor represents a potential zero of the polynomial, which corresponds to a vertical asymptote of the function when it is in the denominator. Factoring makes it easier to determine where a function like \( h(x) \) will have undefined values, leading to discontinuities.
In our exercise, the polynomial \( x^{2}-x-2 \) was factored into \( (x - 2)(x + 1) \), simplifying the identification of discontinuities in the rational function. Each factor represents a potential zero of the polynomial, which corresponds to a vertical asymptote of the function when it is in the denominator. Factoring makes it easier to determine where a function like \( h(x) \) will have undefined values, leading to discontinuities.
Graphing Utilities
Graphing utilities are indispensable tools for visualizing mathematical functions and understanding their properties. These tools range from simple online graphing calculators to complex software programs that can handle a wide array of mathematical tasks. They enable students to plot a graphical representation of a function quickly, assessing its behavior and identifying features like continuity, asymptotes, and intercepts.
When dealing with the function \( h(x) \) in the exercise, a graphing utility provides a visual representation of where the function is continuous and where discontinuities occur without having to evaluate the function at numerous points manually. This visual aid is crucial for grasping the concept of discontinuities and how they manifest on the function's graph.
When dealing with the function \( h(x) \) in the exercise, a graphing utility provides a visual representation of where the function is continuous and where discontinuities occur without having to evaluate the function at numerous points manually. This visual aid is crucial for grasping the concept of discontinuities and how they manifest on the function's graph.
Other exercises in this chapter
Problem 47
Find the point(s), if any, at which the graph of \(f\) has a horizontal tangent. $$ f(x)=\frac{x^{2}}{x-1} $$
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find \(f^{\prime}(x)\). $$ f(x)=x^{4 / 5}+x $$
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Find an equation of the line that is tangent to the graph of \(f\) and parallel to the given line. $$ f(x)=-\frac{1}{4} x^{2} \quad x+y=0 $$
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find the limit $$ \lim _{x \rightarrow-2} \frac{x^{3}+8}{x+2} $$
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