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Chapter 1: Problem 6

Let \(g(x)=-\frac{|x+3|}{x+3}\). a. What is the domain of \(g ?\) b. Use a sequence of values near \(a=-3\) to estimate the value of \(\lim _{x \rightarrow-3} g(x),\) if you think the limit exists. If you think the limit doesn't exist, explain why. c. Use algebra to simplify the expression \(\frac{|x+3|}{x+3}\) and hence work to evaluate \(\lim _{x \rightarrow-3} g(x)\) exactly, if it exists, or to explain how your work shows the limit fails to exist. Discuss how your findings compare to your results in (b). (Hint: \(|a|=a\) whenever \(a \geq 0\), but \(|a|=-a\) whenever \(a<0 .)\) d. True or false: \(g(-3)=-1\). Why? e. True or false: \(-\frac{|x+3|}{x+3}=-1 .\) Why? How is this equality connected to your work above with the function \(g\) ? f. Based on all of your work above, construct an accurate, labeled graph of \(y=g(x)\) on the interval \([-4,-2],\) and write a sentence that explains what you now know about \(\lim _{x \rightarrow-3} g(x)\)

Short Answer

Expert verified
Domain: all x except -3. Limit: Doesn't exist. Graph: y = 1 for x < -3, undefined at x = -3, y = -1 for x > -3.

Step by step solution

01

Determine the Domain of g

The domain of a function is the set of all possible input values (x-values) that it can accept. The given function is \(g(x) = -\frac{|x+3|}{x+3}\). The denominator \(x+3\) cannot be zero because division by zero is undefined. So, \(x+3 eq 0\) or \(x eq -3\). Therefore, the domain of \(g(x)\) is all real numbers except \(x = -3\).
02

Estimate the Limit Using Sequence of Values

To estimate \(\lim_{x \rightarrow -3} g(x)\), choose values of \(x\) close to -3 from both the left and right. For example: When \(x = -3.1, \) \(g(-3.1) = -\frac{|-3.1+3|}{-3.1+3} = -\frac{|-0.1|}{-0.1} = -1\). When \(x = -2.9, \) \(g(-2.9) = -\frac{|-2.9+3|}{-2.9+3} = -\frac{|0.1|}{0.1} = -1\). Since \(|x+3|\) always makes the expression positive and the minus sign in front negates it, \(g(x)\) stays -1 for numbers close to -3. Thus, the estimated value is -1.
03

Simplify the Expression Using Algebra

Let's use the properties of absolute value to simplify: \(\frac{|x+3|}{x+3} = 1\) when \(x + 3 > 0\) and \(\frac{|x+3|}{x+3} = -1\) when \(x + 3 < 0\). Since \( g(x) = -\frac{|x+3|}{x+3} \), for \(x + 3 > 0 \, (i.e., \ x > -3) ====> g(x) = -1 \) And for \(x + 3 < 0 \, (i.e., \ x < -3) ====> g(x) = -(-1) = 1\). From this we see that the function changes behavior drastically around \(x=-3\).
04

Compare Algebraic Findings with Estimated Limit

The previous algebraic deduction shows that \(g(x)\) heads to \(1\) as \(x\) approaches \(-3\) from the left and to \(-1\) as \(x\) moves from the right. So the limit doesn't exist since the two-sided limit isn't the same.
05

Answer the True/False Questions

(d) False: \(g(-3)\) is not defined as \(x = -3\) renders the denominator zero, causing the expression to be undefined. (e) True: whenever \(x + 3 eq 0\) , \( -\frac{|x+3|}{x+3}\) simplifies to -1. This is connected to the function \(g(x)\) since, for any \(x eq -3\), \(g(x) = -1\).
06

Graph the Function on Interval \([-4 to -2]\) & Conclusions

Graph will be a constant positive line of \(y=1\) for \(x < -3\), an undefined point for x=-3, and a constant negative line \( y = -1\) for \( x>-3 \) Hence the limit \(\lim_{x \rightarrow -3} g(x)\) indeed doesn't exist, as the function has a discontinuity at \(-3\).

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Key Concepts

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

Domain of a function
The domain of a function is the set of all possible input values that the function can accept. For the given function \(g(x) = -\frac{|x+3|}{x+3}\), the denominator \(x+3\) must not be zero because division by zero is undefined. Therefore, we set the condition \(x+3 eq 0\) or \(x eq -3\). Thus, the domain of \(g(x)\) is all real numbers except \(x = -3\).

When identifying the domain, always look for values that might cause division by zero, the square root of a negative number, or other undefined operations. In this case, the only problematic value is -3, making the domain all other real numbers.
Limits of functions
Limits are essential in calculus for describing the behavior of a function as it approaches a certain point. To understand \(\lim_{x \rightarrow -3} g(x)\), we examine values of \(x\) close to -3 from both the left and right.

When approaching from the left (\(x < -3\)):
  • If \(x = -3.1\), then \(g(-3.1) = -\frac{|x+3|}{x+3} = -\frac{|-0.1|}{-0.1} = -1\).

When approaching from the right (\(x > -3\)):
  • If \(x = -2.9\), then \(g(-2.9) = -\frac{|x+3|}{x+3} = -\frac{|0.1|}{0.1} = -1\).

In both cases, the function tends to -1, suggesting a one-sided limit is consistent for values close to -3.

Limits help in understanding how a function behaves near a specific point, especially useful in identifying continuity and discontinuity.
Discontinuity in calculus
A function is discontinuous at a point if there is a sudden break or jump. For \(g(x)\), we analyze around \(x = -3\).

By simplifying the expression using properties of the absolute value:
  • For \(x + 3 > 0\) (i.e., \(x > -3\)), \(\frac{|x+3|}{x+3} = 1\) and so \(g(x) = -1\).
  • For \(x + 3 < 0\) (i.e., \(x < -3\)), \(\frac{|x+3|}{x+3} = -1\) and so \(g(x) = 1\).

The function changes from 1 to -1 around \(x = -3\), showcasing a discontinuity where values from left and right of -3 don't match.

The limit \(\lim_{x \rightarrow -3} g(x)\) doesn't exist because the two-sided limits don't align.

Discontinuities are crucial in calculus as they signify points where the function doesn't behave as expected. Identifying them helps in understanding the function鈥檚 behavior better.
Absolute value properties
The absolute value of a number \(a\), denoted \(|a|\), is always non-negative. It is defined as:

  • \(|a| = a\) if \(a \geq 0\)
  • \(|a| = -a\) if \(a < 0\)

Applying these properties, we simplify \(\frac{|x+3|}{x+3}\):
  • For \(x + 3 > 0\): \( |x+3| = x+3\), so \(\frac{|x+3|}{x+3} = 1\).
  • For \(x + 3 < 0\): \( |x+3| = -(x+3)\), so \(\frac{|x+3|}{x+3} = -1\).

Understanding this helps simplify complex expressions and solve limits problems.

This knowledge allows us to break down functions intuitively, seeing changes in behavior based on input values. For \(g(x)\), the minus sign outside the fraction inverts the result, leading to \(-1\) or \(1\) based on the input鈥檚 sign, and helps explain the function's discontinuity at \(x=-3\).

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Most popular questions from this chapter

A certain function \(y=p(x)\) has its local linearization at \(a=3\) given by \(L(x)=-2 x+5\). a. What are the values of \(p(3)\) and \(p^{\prime}(3) ?\) Why? b. Estimate the value of \(p(2.79)\). c. Suppose that \(p^{\prime \prime}(3)=0\) and you know that \(p^{\prime \prime}(x)<0\) for \(x<3\). Is your estimate in (b) too large or too small? d. Suppose that \(p^{\prime \prime}(x)>0\) for \(x>3\). Use this fact and the additional information above to sketch an accurate graph of \(y=p(x)\) near \(x=3\). Include a sketch of \(y=L(x)\) in your work.

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For each of the following prompts, sketch a graph on the provided axes of a function that has the stated properties. a. \(y=f(x)\) such that - \(f(-2)=2\) and \(\lim _{x \rightarrow-2} f(x)=1\) \- \(f(-1)=3\) and \(\lim _{x \rightarrow-1} f(x)=3\) \- \(f(1)\) is not defined and \(\lim _{x \rightarrow 1} f(x)=0\) \- \(f(2)=1\) and \(\lim _{x \rightarrow 2} f(x)\) does not exist. b. \(y=g(x)\) such that \- \(\text { - } g(-2)=3, g(-1)=-1, g(1)=-2, \text { and } g(2)=3\) \- At \(x=-2,-1,1\) and \(2, g\) has a limit, and its limit equals the value of the function at that point. \- \(g(0)\) is not defined and \(\lim _{x \rightarrow 0} g(x)\) does not exist.

For the function $$f(x)=\left\\{\begin{array}{ll} 2 x-5, & 0 \leq x<3 \\ 4, & x=3 \\ x^{2}-6 x+10, & 3

A diver leaps from a 3 meter springboard. His feet leave the board at time \(t=0\), he reaches his maximum height of \(4.5 \mathrm{~m}\) at \(t=1.1\) seconds, and enters the water at \(t=2.45\). Once in the water, the diver coasts to the bottom of the pool (depth \(3.5 \mathrm{~m}\) ), touches bottom at \(t=7\), rests for one second, and then pushes off the bottom. From there he coasts to the surface, and takes his first breath at \(t=13\). a. Let \(s(t)\) denote the function that gives the height of the diver's feet (in meters) above the water at time \(t\). (Note that the "height" of the bottom of the pool is -3.5 meters.) Sketch a carefully labeled graph of \(s(t)\) on the provided axes in Figure 1.1.5. Include scale and units on the vertical axis. Be as detailed as possible. b. Based on your graph in (a), what is the average velocity of the diver between \(t=2.45\) and \(t=7 ?\) Is his average velocity the same on every time interval within [2.45,7]\(?\) c. Let the function \(v(t)\) represent the instantaneous vertical velocity of the diver at time \(t\) (i.e. the speed at which the height function \(s(t)\) is changing; note that velocity in the upward direction is positive, while the velocity of a falling object is negative). Based on your understanding of the diver's behavior, as well as your graph of the position function, sketch a carefully labeled graph of \(v(t)\) on the axes provided in Figure \(1.1 .6 .\) Include scale and units on the vertical axis. Write several sentences that explain how you constructed your graph, discussing when you expect \(v(t)\) to be zero, positive, negative, relatively large, and relatively small. d. Is there a connection between the two graphs that you can describe? What can you say about the velocity graph when the height function is increasing? decreasing? Make as many observations as you can.
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