Problem 6 - Active Calculus

Question

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 \(\lim _{x \rightarrow-3} g(x)\)

Step-by-Step Solution

Answer

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 1: 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 \(x = -3\).

Step 2: 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.

Step 3: 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 \(x=-3\).

Step 4: 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.

Step 5: 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 \(x eq -3\), \(g(x) = -1\).

Step 6: 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 \(-3\).

Key Concepts

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’s 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’s sign, and helps explain the function's discontinuity at \(x=-3\).