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Determine at which of the points \(-1,1,\) and 2 the given function \(f\) is continuous. $$ f(x)=\left\\{\begin{array}{cl} \left(x^{3}-x\right) /\left(x^{2}-1\right) & \text { if } x<-1 \\ 2 x+1 & \text { if }-1 \leq x \leq 2 \\ \left(x^{2}-4\right) /(x-2) & \text { if } 2

Short Answer

Expert verified
The function is only continuous at \( x = 1 \).

Step by step solution

01

Evaluate Continuity at x = -1

To check continuity at \( x = -1 \), we verify if the following three conditions hold: (1) \( f(-1) \) is defined, (2) \( \lim_{{x \to -1^-}} f(x) = \lim_{{x \to -1^+}} f(x) = f(-1) \). Calculate \( f(-1) \) using the second piece: \( f(-1) = 2(-1) + 1 = -1 \). Now, test assuming the limit from the left side, \( \lim_{{x \to -1^-}} = \lim_{{x \to -1^-}} \frac{x^3 - x}{x^2 - 1} \). The function undefined near \(x = -1\), hence \( \lim_{{x \to -1^-}} f(x) \) does not match \(-1\). Thus, the function is not continuous at \(x = -1\).
02

Evaluate Continuity at x = 1

For continuity at \( x = 1 \), ensure: (1) \( f(1) \) is defined; (2) \( \lim_{{x \to 1^-}} f(x) = \lim_{{x \to 1^+}} f(x) = f(1) \). Calculate \( f(1) \) with \( 2x+1: f(1) = 2(1) + 1 = 3 \). Both limits from the left and right of 1 (both in the form \( 2x + 1 \)) are 3. Therefore, the function is continuous at \( x = 1 \).
03

Evaluate Continuity at x = 2

Check conditions (1) \( f(2) \) is defined, and (2) \( \lim_{{x \to 2^-}} f(x) = \lim_{{x \to 2^+}} f(x) = f(2) \). Calculate \( f(2) \) using the middle piece \( 2(2) + 1 = 5 \). For \( \lim_{{x \to 2^-}} \), \( f(x) = 2x + 1 \), which evaluates to 5. Calculate \( \lim_{{x \to 2^+}} \): the provided third piece, simplifies to \( x+2 \) using \( x^2-4 = (x-2)(x+2) \), leading the limit to 4. As \( \lim_{{x \to 2^+}} eq \lim_{{x \to 2^-}} \), the function is not continuous at \( x = 2 \).

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Key Concepts

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

Piecewise Functions
Piecewise functions are mathematical expressions defined by multiple sub-functions, each applicable to a specific interval of the main function's domain. These functions display different behaviors depending on the range of input values they receive.
For example, in the given function:
  • If \( x < -1 \), the expression is \( \frac{x^3 - x}{x^2 - 1} \).
  • If \(-1 \leq x \leq 2\), the expression is \( 2x + 1 \).
  • If \( x > 2 \), the expression is \( \frac{x^2 - 4}{x - 2} \).
Switching to different formulas regarding the intervals results in the piecewise nature. The critical task is how to manage the transitions between these sub-functions smoothly.
This can be particularly delicate at points where the piece transitions, like at boundaries \( x = -1 \), \( x = 1 \), or \( x = 2 \). Ensuring continuity involves verifying specific conditions, which can turn these math pieces into smoothly traversed parts of a function.
Limit Evaluation
Limit evaluation in calculus is crucial when determining the behavior of functions as they approach certain points or infinity. It examines the values a function tends towards as the variable gets closer to a particular point.
The steps to evaluate a limit often involve substitution or algebraic manipulation to find an expression that works for the entire domain except the point in question. This can require different techniques for piecewise functions.
For example, to find \( \lim_{{x \to -1^-}} \) of \( \frac{x^3 - x}{x^2 - 1} \), one might simplify or rearrange the expression to handle direct computation or decompose it to remove undefined operations.
Limit analysis helps determine whether a function is continuous at a point by checking if the left-hand and right-hand limits equal and match the function's value at that point (if defined). Thus, mastering limit evaluation is key to grasping function behavior near those crucial transition points.
Discontinuity Points
Discontinuity points are specific input values where a function isn't continuous. At these points, the function fails the basic continuity checks, which include:
  • The function is defined at the point of interest.
  • The left-hand limit equals the right-hand limit at that point.
  • Both limits (from either side) are equal to the value of the function at the point.
When even one of these conditions isn't met, the function is discontinuous at that location. Discontinuity can be of various types:
  • Removable: If the function can be redefined at that point to "fix" the discontinuity.
  • Jump: If the left and right-hand limits exist but aren't equal.
  • Infinite: If the function grows indefinitely at that point.
In our example, \( x = -1 \) and \( x = 2 \) are discontinuity points. At \( x = -1 \), the function isn't defined from the left side. At \( x = 2 \), a jump discontinuity exists because the limits from either side don't equal each other, nor do they match \( f(2) \). Understanding these classifications helps in figuring out how a function behaves and what kind of corrections, if any, can bring about continuity.

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Most popular questions from this chapter

Plot \(y=\exp (x)\) for \(0 \leq x \leq 2\). Let \(P(c)\) denote the point \((c, \exp (c))\) on the graph. The purpose of this exercise is to graphically explore the relationship between \(\exp (c)\) and the slope of the tangent line at \(P(c) .\) For \(c=1 / 2,1\), and \(3 / 2,\) calculate the slope \(m(c)\) of the secant line that passes through the pair of points \(P(c-0.001)\) and \(P(c+0.001) .\) For each \(c,\) calculate \(|\exp (c)-m(c)|\) to see that \(m(c)\) is a good approximation of \(\exp (c) .\) Add the three secant lines to your viewing window. For each of \(c=1 / 2,1,\) and \(3 / 2,\) add to the viewing window the line through \(P(c)\) with slope \(\exp (c)\). As we will see in Chapter \(3,\) these are the tangent lines at \(P(1 / 2), P(1)\) and \(P(3 / 2)\). It is likely that they cannot be distinguished from the secant lines in your plot.

If an amount \(A\) is to be received at time \(T\) in the future, then the present value of that payment is the amount \(P_{0}\) that, if deposited immediately with the current interest rate locked in, will grow to \(A\) by time \(T\) under continuous compounding. The present value of an income stream is the sum of the present values of each future payment. In each of Exercise \(82-85,\) calculate the present value of the specified income stream. Every exponential function can be expressed by means of the natural exponential function. To be specific, for any fixed \(a>0,\) there is a constant \(k\) such that \(a^{x}=e^{k x}\) for every \(x\). Express \(k\) in terms of \(a\).

A function \(f\) and a point \(c\) not in the domain of \(f\) are given. Analyze \(\lim _{x \rightarrow c} f(x)\) as follows. a. Evaluate \(f\left(c-1 / 10^{n}\right)\) and \(f\left(c+1 / 10^{n}\right)\) for \(n=2,3,4\). b. Formulate a guess for the value \(\ell=\lim _{x \rightarrow c} f(x)\). c. Find a value \(\delta\) such that \(f(x)\) is within 0.01 of \(\ell\) for every \(x\) that is within \(\delta\) of \(c\). d. Graph \(y=f(x)\) for \(x\) in \(\left[c-1 / 10^{4}, c\right) \cup\left(c, c+1 / 10^{4}\right]\) to verify visually that the limit of \(f\) at \(c\) exists. $$ f(x)=\frac{\sin (x)}{x-\pi}, c=\pi $$

An assertion is made about a function \(f\) that is defined on a closed, bounded interval. If the statement is true, explain why. Otherwise, sketch a function \(f\) that shows it is false. (Note: \(|f|\) is defined by \(|f|(x)=|f(x)| .)\) If \(f\) is continuous, then \(f\) is continuous.

In Exercises \(67-73,\) use algebraic manipulation (as in Example 5 ) to evaluate the limit. $$ \lim _{h \rightarrow 0} \frac{h}{\sqrt{1+2 h}-1} $$

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