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Chapter 2: Problem 18

Find the points of discontinuity, if any. $$f(x)=\frac{5}{x}+\frac{2 x}{x+4}$$

Short Answer

Expert verified
The points of discontinuity are \( x = 0 \) and \( x = -4 \).

Step by step solution

01

Identify Points of Undefined Expression

A function is discontinuous at points where it is undefined. Check where \( f(x) = \frac{5}{x} + \frac{2x}{x+4} \) is undefined by setting the denominators to zero. Solve \( x = 0 \) and \( x + 4 = 0 \). The solutions, \( x = 0 \) and \( x = -4 \), are potential points of discontinuity.
02

Analyze Each Point of Discontinuity

We have identified \( x = 0 \) and \( x = -4 \) as potential points of discontinuity. Both points make the denominators of the function zero, which means a division by zero occurs and the function is undefined at these points.

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Key Concepts

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

Undefined Function
An undefined function occurs in mathematical expressions when certain operations cannot be executed usually due to a specific condition. In functions like \( f(x) = \frac{5}{x} + \frac{2x}{x+4} \), the presence of variables in the denominator is a common source of undefined expressions. If the denominator equals zero, the function cannot compute a real number, leading to an undefined result. This is because division by zero is mathematically impossible.
  • Identify when a function is undefined: Look for any denominators set equal to zero.
  • Evaluate at those points: Substitute the values back into the denominator to verify if it causes an undefined scenario.
In our original problem, \( x = 0 \) and \( x = -4 \) caused the denominator of each term to be zero, hence making the function undefined at these points.
Division by Zero
Division by zero is a critical concept in mathematics that one must avoid, as it disrupts the basic rules of arithmetic. Normally, division involves splitting a number into parts. However, when dividing by zero, there鈥檚 no logical way to distribute quantities into zero parts. This leads to a result that is not defined.
  • Mathematically invalid: Division by zero doesn鈥檛 produce a number or a defined value.
  • Indicator of discontinuity: Points where division by zero happens are usually where the function has breaks or is not continuous.
In the function \( f(x) = \frac{5}{x} + \frac{2x}{x+4} \), attempting to evaluate at \( x = 0 \) or \( x = -4 \) leads to divisions by zero, confirming the discontinuity at these points.
Points of Discontinuity
Points of discontinuity in a function are specific values of the variable where the function does not remain continuous. A continuous function can be drawn without lifting a pencil from the paper, while a discontinuity indicates a break or gap.
  • Types of discontinuity: Can be removable or non-removable. Removable when a simple algebraic adjustment can make the function continuous, otherwise non-removable.
  • Identification: Usually associated with undefined values or division by zero in the function.
In the given exercise, the points \( x = 0 \) and \( x = -4 \) mark discontinuities because these are where the function is undefined due to division by zero. Such points mean you cannot graph the function smoothly through these values, resulting in noticeable breaks in its curve.

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Most popular questions from this chapter

Find the limits. $$\lim _{x \rightarrow 3} \frac{x}{x-3}$$

Find a value for the constant \(k\) that makes $$f(x)=\left\\{\begin{array}{ll}\frac{\sin 3 x}{x}, & x \neq 0 \\\k, & x=0\end{array}\right.$$ continuous at \(x=0\)

Find the limits. $$\lim _{x \rightarrow 1^{+}} \frac{x^{4}-1}{x-1}$$

(a) Use a graphing utility to generate the graph of the function \(f(x)=(x+3) /\left(2 x^{2}+5 x-3\right),\) and then use the graph to make a conjecture about the number and location of all discontinuities. (b) Check your conjecture by factoring the denominator.

Express the limit as an equivalent limit in which \(x \rightarrow 0^{+}\) or \(x \rightarrow 0^{-}\), as appropriate. I You need not evaluate the limit. \(]\) (a) \(\lim _{x \rightarrow+\infty} \frac{\cos (\pi / x)}{\pi / x}\) (b) \(\lim _{x \rightarrow+\infty} \frac{x}{1+x}\) (c) \(\lim _{x \rightarrow-\infty}(1+2 x)^{1 / x}\)
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