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Chapter 1: Problem 4

Find the value(s) of \(x\) at which each function is discontinuous. a. \(f(x)=\frac{9-x^{2}}{x-3}\) b. \(g(x)=\frac{7 x-4}{x}\) c. \(h(x)=\frac{x^{2}+1}{x^{3}}\) d. \(f(x)=\frac{x-4}{x^{2}-9}\) e. \(g(x)=\frac{13 x}{x^{2}+x-6}\) f. \(h(x)=\left\\{\begin{array}{r}-x, \text { if } x \leq 3 \\ 1-x, \text { if } x>3\end{array}\right.\)

Short Answer

Expert verified
a. 3; b. 0; c. 0; d. 3, -3; e. -3, 2; f. No discontinuity.

Step by step solution

01

Identify Points of Discontinuity for Function a

The function given is \(f(x)=\frac{9-x^{2}}{x-3}\). To find where it is discontinuous, determine when the denominator equals zero:\[ x-3 = 0 \] Solving this gives \(x = 3\). At \(x = 3\), the function is undefined, thus it is a point of discontinuity.
02

Identify Points of Discontinuity for Function b

The function is \(g(x)=\frac{7x-4}{x}\). A function with a denominator of zero is discontinuous. Solve the equation for zero:\[ x = 0 \] Therefore, the function is discontinuous at \(x = 0\).
03

Identify Points of Discontinuity for Function c

Given \(h(x)=\frac{x^{2}+1}{x^{3}}\), find points where the denominator is zero:\[ x^{3} = 0 \] Solving this gives \(x = 0\). Thus, \(h(x)\) is discontinuous at \(x = 0\).
04

Identify Points of Discontinuity for Function d

The function provided is \(f(x)=\frac{x-4}{x^{2}-9}\). Find when the denominator is zero:\[ x^{2} - 9 = 0 \] This factors to \((x-3)(x+3)=0\), so \(x=3\) and \(x=-3\). The function is discontinuous at \(x = 3\) and \(x = -3\).
05

Identify Points of Discontinuity for Function e

The function \(g(x)=\frac{13x}{x^{2}+x-6}\) is discontinuous where its denominator is zero. Factor the denominator:\[ x^{2}+x-6 = (x+3)(x-2) \] Setting this equal to zero, we get \(x = -3\) and \(x = 2\). These are the points of discontinuity.
06

Identify Points of Discontinuity for Function f

The piecewise function \(h(x)=\left\{ \begin{array}{rr} -x, & \text{if } x \leq 3 \ 1-x, & \text{if } x>3 \end{array} \right.\) is often discontinuous at its boundaries. Check continuity at \(x = 3\):\[-3 = 1-3 \] Since both expressions equal -3, the function is continuous at \(x = 3\), meaning there are no points of discontinuity.

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Key Concepts

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

Rational Functions
Rational functions are fractions that have polynomials in their numerator and denominator, like \( f(x) = \frac{9-x^2}{x-3} \). These types of functions present particular challenges when the denominator becomes zero, because division by zero is undefined in mathematics.
  • When the denominator of a rational function equals zero, it creates a point where the function is discontinuous. So, to find these points, we set the denominator equal to zero and solve for \(x\).
  • For instance, in \(f(x) = \frac{9-x^2}{x-3}\), setting \(x-3 = 0\) reveals that \(x=3\) is where the function is undefined, causing discontinuity.
Understanding the points of discontinuity is crucial in dealing with rational functions. It not only affects the domain of the function but also leads to the concept of vertical asymptotes, which we鈥檒l explore further.
Piecewise Functions
Piecewise functions are defined by multiple sub-functions, each with its own domain. A classic example is \( h(x) = \left\{\begin{array}{r} -x, \text{ if } x \leq 3 \ 1-x, \text{ if } x>3 \end{array}\right. \). This means the function rule changes depending on the input value.
  • To check for discontinuity, focus on each "border" where the definition of the function switches.
  • Evaluate the function's behavior right at the transition points. This often involves matching the function's value as it "moves" from one part to another.
In our example, at \(x = 3\), both sub-functions equate to \(-3\), ensuring there is no discontinuity at this point. Thus, while piecewise functions might seem complex, careful analysis at transition points simplifies things a lot.
Asymptotes
Asymptotes are lines that a graph approaches but never touches or crosses. They are significant in understanding rational functions, where vertical asymptotes correspond to points of discontinuity. These occur when the denominator equals zero at a point where the numerator does not also equal zero.
  • Vertical asymptotes are suggested at discontinuities where the limit of the function as it approaches the point is infinite.
  • For example, in \( g(x) = \frac{7x-4}{x} \), \( x = 0 \) is a vertical asymptote since the function approaches \( \pm \infty \) near it, making it discontinuous.
Recognizing asymptotes helps in sketching graphs and understanding function behavior far from these critical points. They serve as invisible boundaries reflected in the graph's shape.
Limits in Calculus
Calculus is all about understanding limits, essential in analyzing function behavior near discontinuities. The limit is the value a function approaches as the input nears a certain point, even if it doesn't reach that point.
  • Finding discontinuities often involves calculating the limits of functions at specific points. It helps determine whether functions approach the same value from both sides.
  • Consider \(f(x) = \frac{x-4}{x^2-9}\). At points \(x=3\) and \(x=-3\), finding limits reveals whether the function "jumps" or approaches infinity.
Limits allow us to make sense of potentially undefined values and help predict behavior, which is why they're a cornerstone of calculus. They bridge discrete mathematical descriptions with continuous function behavior.

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Most popular questions from this chapter

A medicine is administered to a patient. The amount of medicine \(M,\) in milligrams, in \(1 \mathrm{mL}\) of the patient's blood, \(t\) hours after the injection, is \(M(t)=-\frac{1}{3} t^{2}+t,\) where \(0 \leq t \leq 3\) a. Find the rate of change in the amount \(M, 2 \mathrm{h}\) after the injection. b. What is the significance of the fact that your answer is negative?

Suppose that a foreign-language student has learned \(N(t)=20 t-t^{2}\) vocabulary terms after \(t\) hours of uninterrupted study, where \(0 \leq t \leq 10\) a. How many terms are learned between time \(t=2 \mathrm{h}\) and \(t=3 \mathrm{h} ?\) b. What is the rate, in terms per hour, at which the student is learning at time \(t=2 \mathrm{h} ?\)

Consider the following function: \(g(x)=\left\\{\begin{array}{c}\frac{x|x-1|}{x-1}, \text { if } x \neq 1 \\ 0, \text { if } x=1\end{array}\right.\) a. Evaluate \(\lim _{x \rightarrow 1^{+}} g(x)\) and \(\lim _{x \rightarrow 1^{-}} g(x),\) and then determine whether \(\lim _{x \rightarrow 1} g(x)\) exists. b. Sketch the graph of \(g(x),\) and identify any points of discontinuity.

Evaluate the limit of each indeterminate quotient. a. \(\lim _{x \rightarrow 2} \frac{4-x^{2}}{2-x}\) b. \(\lim _{x \rightarrow-1} \frac{2 x^{2}+5 x+3}{x+1}\) c. \(\lim _{x \rightarrow 3} \frac{x^{3}-27}{x-3}\) d. \(\lim _{x \rightarrow 0} \frac{2-\sqrt{4+x}}{x}\) e. \(\lim _{x \rightarrow 4} \frac{\sqrt{x}-2}{x-4}\) f. \(\lim _{x \rightarrow 0} \frac{\sqrt{7-x}-\sqrt{7+x}}{x}\)

Find the points on the graph of \(y=\frac{1}{3} x^{3}-5 x-\frac{4}{x}\) at which the tangent is horizontal.
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