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Find the points of discontinuity, and determine whether the discontinuities are removable. (a) \(f(x)=\frac{x^{2}-4}{x^{3}-8}\) (b) \(f(x)=\left\\{\begin{array}{ll}2 x-3, & x \leq 2 \\ x^{2}, & x>2\end{array}\right.\) (c) \(f(x)=\left\\{\begin{array}{ll}3 x^{2}+5, & x \neq 1 \\ 6, & x=1\end{array}\right.\)

Short Answer

Expert verified
(a) Removable at \( x=2 \), (b) Non-removable at \( x=2 \), (c) Non-removable at \( x=1 \).

Step by step solution

01

Identify Discontinuities for (a)

For the function \( f(x) = \frac{x^2 - 4}{x^3 - 8} \), check the denominator to find points of discontinuity. Set \( x^3 - 8 = 0 \), and solve for \( x \): \( x^3 = 8 \), so \( x = 2 \). The function is potentially discontinuous at \( x = 2 \).
02

Determine Removability for (a)

Factor both the numerator and the denominator of \( \frac{x^2 - 4}{x^3 - 8} \). The numerator \( x^2 - 4 \) factors to \( (x - 2)(x + 2) \), and the denominator \( x^3 - 8 \) factors to \( (x - 2)(x^2 + 2x + 4) \). The \( (x - 2) \) terms cancel each other, resulting in \( \frac{x+2}{x^2+2x+4} \). Since the discontinuity is removable, the new function is continuous at \( x = 2 \).
03

Identify Discontinuities for (b)

For the piecewise function \( f(x)= \begin{cases} 2x - 3, & x \leq 2 \ x^2, & x > 2 \end{cases} \), a potential discontinuity exists at \( x = 2 \), where the definition changes.
04

Determine Removability for (b)

Check the limits as \( x \to 2 \) from the left and right: \( \lim_{x \to 2^-} (2x - 3) = 2(2) - 3 = 1 \) and \( \lim_{x \to 2^+} x^2 = 2^2 = 4 \). Since these two limits are not equal, the function has a non-removable discontinuity at \( x = 2 \).
05

Identify Discontinuities for (c)

The piecewise function \( f(x)= \begin{cases} 3x^2 + 5, & x eq 1 \ 6, & x = 1 \end{cases} \) is potentially discontinuous at \( x = 1 \).
06

Determine Removability for (c)

Evaluate the limits on both sides of 1: \( \lim_{x \to 1} (3x^2 + 5) = 3(1)^2 + 5 = 8 \). Since \( f(1) = 6 \), the limit does not equal \( f(1) \), indicating a non-removable discontinuity at \( x = 1 \).

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Key Concepts

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

Removable Discontinuity
When we talk about a removable discontinuity, we're referring to a point on a function's graph where there's a hole. This happens when we can "fix" the function at that point by redefining it. It's like a small interruption that can easily be patched up.

Take, for instance, the function \( f(x) = \frac{x^2 - 4}{x^3 - 8} \). We found that at \( x = 2 \), the function is initially discontinuous. However, since we managed to factor and cancel \( (x-2) \) in the numerator and denominator, the discontinuity at this point was removable. We are left with the continuous function \( \frac{x+2}{x^2+2x+4} \) at \( x = 2 \).

In this case, the removable discontinuity means we "cancel out the problem", so the function continues smoothly as if the interruption never existed. This is a key feature of removable discontinuities—it's all about simple adjustments.
Non-Removable Discontinuity
Non-removable discontinuities are a bit different. These occur where no simple tweaking can re-establish the continuity of the function. They often arise in step or jump form.

Let's look at a piecewise function like \( f(x)= \begin{cases} 2x - 3, & x \leq 2 \ x^2, & x > 2 \end{cases} \). Here, there's a break at \( x = 2 \). The limit from the left side is 1, whereas from the right it is 4. Different limits mean a definite gap; thus, this is a non-removable discontinuity. No matter how we redefine \( f(2) \), we can't fill this gap seamlessly.

Another example is the function \( f(x)= \begin{cases} 3x^2 + 5, & x eq 1 \ 6, & x = 1 \end{cases} \). At \( x=1 \), the graph doesn't smoothly continue, because the function value (6) and the limit (8) are not the same. This separation confirms it's a non-removable discontinuity.
Piecewise Functions
Piecewise functions can be thought of as mathematical patchwork quilts. They stitch different function rules over specified intervals.

In a piecewise function, each "piece" functions according to its own rule in its domain. For example, consider the function \( f(x)= \begin{cases} 2x - 3, & x \leq 2 \ x^2, & x > 2 \end{cases} \). This is composed of two parts, a linear segment when \( x \leq 2 \), and a quadratic segment when \( x > 2 \). The potential for discontinuity arises at the juncture where these segments meet, in this case, \( x = 2 \).

Another example, \( f(x)= \begin{cases} 3x^2 + 5, & x eq 1 \ 6, & x = 1 \end{cases} \), comprises a continuous quadratic expression applied universally except at a specific point \( x = 1 \).

These functions can present both removable and non-removable discontinuities, making it essential to evaluate limits and endpoints to fully understand their behavior across specified intervals.

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Most popular questions from this chapter

In each part, find the largest open interval, centered at the point \(x=1,\) such that for each point \(x\) in the interval the value of \(f(x)=1 /(x-1)^{2}\) is greater than \(M\) (a) \(M=10\) (b) \(M=1000\) (c) \(M=100.000\)

(a) Find the smallest positive number \(N\) such that for each point \(x\) in the interval \((N,+\infty),\) the value of the function \(f(x)=1 / x^{2}\) is within 0.1 unit of \(L=0\) (b) Find the smallest positive number \(N\) such that for each point \(x\) in the interval \((N,+\infty),\) the value of \(f(x)=x /(x+1)\) is within 0.01 unit of \(L=1\) (c) Find the largest negative number \(N\) such that for each point \(x\) in the interval \((-\infty, N)\). the value of the function \(f(x)=1 / x^{3}\) is within 0.001 unit of \(L=0\) (d) Find the largest negative number \(N\) such that for each point \(x\) in the interval \((-\infty, N)\), the value of the function \(f(x)=x /(x+1)\) is within 0.01 unit of \(L=1\)

Find the limits. $$\lim _{h \rightarrow 0} \frac{h}{\tan h}$$

(i) Approximate the \(y\) -coordinates of all horizontal asymptotes of \(y=f(x)\) by evaluating \(f\) at the points \(\pm 10 . \pm 100 . \pm 1000 .\pm 100.000,\) and \(\pm 100.000 .000\) (ii) Confirm your conclusions by graphing \(y=f(x)\) over an appropriate interval. (iii) If you have a CAS, then use it to find the horizontal asymptotes. (a) \(f(x)=\frac{2 x+3}{x+4}\) (b) \(f(x)=\left(1+\frac{3}{x}\right)^{x}\) (c) \(f(x)=\frac{x^{2}+1}{x+1}\)

Find the points of discontinuity, if any. $$f(x)=\frac{x+3}{\left|x^{2}+3 x\right|}$$

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