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Chapter 1: Problem 86

Graph the function \(f\) given by $$ f(x)=\left\\{\begin{array}{ll} -3, & \text { for } x=-2 \\ x^{2}, & \text { for } x \neq-2 \end{array}\right. $$ Use GRAPH and TRACE to find each of the following limits. When necessary, state that the limit does not exist. a) \(\lim _{x \rightarrow-2^{+}} f(x)\) b) \(\lim _{x \rightarrow-2^{-}} f(x)\) c) \(\lim _{x \rightarrow-2} f(x)\) d) \(\lim _{x \rightarrow 2^{+}} f(x)\) e) \(\lim _{x \rightarrow 2^{-}} f(x)\) f) Does \(\lim _{x \rightarrow-2} f(x)=f(-2)\) ? g) Does \(\lim _{x \rightarrow 2} f(x)=f(2)\) ?

Short Answer

Expert verified
a) 4, b) 4, c) 4, d) 4, e) 4, f) No, g) Yes.

Step by step solution

01

Analyzing the Function

The function is a piecewise function. For \( x = -2 \), the function takes a fixed value \( f(x) = -3 \). For \( x eq -2 \), the function behaves like \( f(x) = x^2 \).
02

Finding the Right-Side Limit at x = -2

Consider the function when approaching \( x = -2 \) from the right (i.e., for values greater than -2). In this case, the function behaves as \( f(x) = x^2 \). Hence, \( \lim_{x \to -2^{+}} f(x) = (-2)^2 = 4 \).
03

Finding the Left-Side Limit at x = -2

Consider the function when approaching \( x = -2 \) from the left (i.e., for values less than -2). Here, the function also follows \( f(x) = x^2 \). Therefore, \( \lim_{x \to -2^{-}} f(x) = (-2)^2 = 4 \).
04

Determining the Limit at x = -2

Since the function behavior follows \( x^2 \) from both sides for \( x eq -2 \), and both the right and left side limits at \( x = -2 \) are 4, we find \( \lim_{x \to -2} f(x) = 4 \).
05

Finding the Right-Side Limit at x = 2

As \( x \to 2^{+} \), the function prefers the \( f(x) = x^2 \) behavior, thus \( \lim_{x \to 2^{+}} f(x) = 2^2 = 4 \).
06

Finding the Left-Side Limit at x = 2

Similarly, as \( x \to 2^{-} \), the function follows \( f(x) = x^2 \), which means \( \lim_{x \to 2^{-}} f(x) = 2^2 = 4 \).
07

Evaluating the Limit at x = 2

Since the limits from both sides as \( x \to 2 \) are equal to 4, we have \( \lim_{x \to 2} f(x) = 4 \).
08

Comparing Limit and Function Value at x = -2

Compared \( \lim_{x \to -2} f(x) = 4 \) with \( f(-2) = -3 \). Since these are different, \( \lim_{x \to -2} f(x) eq f(-2) \).
09

Comparing Limit and Function Value at x = 2

Here, \( \lim_{x \to 2} f(x) = 4 \) and \( f(2) = 2^2 = 4 \). Thus, \( \lim_{x \to 2} f(x) = f(2) \).

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Key Concepts

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

Limits of Functions
When studying piecewise functions, limits help us understand how the function behaves as it approaches specific points from either side. For the given function, we seek to determine the limit at certain points, particularly at \(-2\) and \(2\). The limit of a function at a point describes the value the function approaches as the input comes closer to the variable's specified value, irrespective of whether it equals that value.
Let's focus on the limits near \(x = -2\):
  • The limit \( \lim_{x \rightarrow -2^{+}} f(x) \) is the value that \( f(x) \) approaches as \( x \) approaches \(-2\) from the right (numbers greater than \(-2\)), which results in \(4\), following the quadratic component \(x^2\).
  • Conversely, \( \lim_{x \rightarrow -2^{-}} f(x) \) refers to the value the function approaches from the left (numbers less than \(-2\)), also evaluating to \(4\).
  • Both these limits establish the overall limit at \( \lim_{x \rightarrow -2} f(x) = 4 \). This underlines the convergence behavior of the function at the point, though the function's explicit value at \\(-2\) differs.
Understanding limits in such scenarios enables us to assess the function's behavior from various approaches, offering deeper insight into its nature.
Continuity
In mathematics, continuity of a function at a point means that the value of the function and the limit of the function as it approaches that point are the same. A function can be continuous over a range, meaning there are no interruptions, holes, or jumps in the graph. For our function:
  • Continuity at \(x = -2\) is problematic. We've identified that \(\lim_{x \rightarrow -2} f(x) = 4\) but \(f(-2) = -3\). Hence, there's a break of continuity at \(-2\).
  • On the other hand, for \(x = 2\), the limit as \(x\) approaches 2 from both sides aligns with the actual value of the function: \(\lim_{x \rightarrow 2} f(x) = 4\), and \(f(2) = 4\). Thus, the function is continuous at \(x = 2\).
Identifying these points of disparity helps understanding where functions have potential disruptions, often crucial for solving real-world problems where continuity indicates stability and predictability.
Graphing Piecewise Functions
Graphing piecewise functions involves plotting different expressions according to specified intervals. In this specific task, the function combines a constant section with a variable section that follows a familiar parabolic shape.
  • At \(x = -2\), a stark value of \(-3\) presents itself, marked usually by a point differently colored or sized to denote the unique element.
  • For all other values of \(x\), the function graph resembles a standard parabola \(y = x^2\), a graceful upward curve representing its natural progression over the rest of the domain.
  • The visual break at \(x = -2\) can be indicated through an open circle, reflecting the fact that the quadratic behavior does not govern that particular point, emphasizing the non-continuous aspect of the function there.
Learning to graph piecewise functions effectively allows students to visualize abstract mathematical concepts, providing an intuitive grasp of functional behaviors and interactions between different mathematical expressions.

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Most popular questions from this chapter

Population growth rate. In \(t\) years, the population of Kingsville grows from 100,000 to a size \(P\) given by \(P(t)=100,000+2000 t^{2}\) a) Find the growth rate, \(d P / d t\). b) Find the population after 10 yr. c) Find the growth rate at \(t=10\). d) Explain the meaning of your answer to part (c).

Let \(f\) and \(g\) be differentiable over an open interval containing \(x=a\). If $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{0}{0} \quad \text { or } \quad \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{\pm \infty}{\pm \infty} $$ and if \(\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right)\) exists, then $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right) $$ The forms \(0 / 0\) and \(\pm \infty / \pm \infty\) are said to be indeterminate. In such cases, the limit may exist, and l'H么pital's Rule offers a way to find the limit using differentiation. For example, in Example 1 of Section \(1.1,\) we showed that $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=2 $$ Since, for \(x=1,\) we have \(\left(x^{2}-1\right)(x-1)=0 / 0,\) we differentiate the numerator and denominator separately, and reevaluate the limit: $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=\lim _{x \rightarrow 1}\left(\frac{2 x}{1}\right)=2 $$ Use this method to find the following limits. Be sure to check that the initial substitution results in an indeterminate form. $$ \lim _{x \rightarrow 5}\left(\frac{x^{2}-25}{2 x-10}\right) $$

Find the simplified difference quotient for each function listed. $$ f(x)=\frac{1}{1-x} $$

Graph \(s,\) v, and a over the given interval. Then use the graphs to determine the point(s) at which the velocity switches from increasing to decreasing or from decreasing to increasing. $$ s(t)=t^{4}+t^{3}-4 t^{2}-2 t+4 ; \quad[-3,3] $$

Find the simplified difference quotient for each function listed. $$ f(x)=a x^{2}+b x+c $$
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