Problem 54
Question
Graph each function and then find the specified limits. When necessary, state that the limit does not exist. $$ g(x)=|x|+1 ; \text { find } \lim _{x \rightarrow-3} g(x) \text { and } \lim _{x \rightarrow 0} g(x). $$
Step-by-Step Solution
Verified Answer
\( \lim_{x \rightarrow -3} g(x) = 4 \) and \( \lim_{x \rightarrow 0} g(x) = 1 \).
1Step 1: Understanding the Function
The function given is \( g(x) = |x| + 1 \). This function describes the absolute value of \( x \), shifted upward by 1 unit. The graph of \( |x| \) is V-shaped, centered at the origin, and adding 1 shifts it upwards.
2Step 2: Graphing the Function
To graph \( g(x) = |x| + 1 \), draw a V-shaped graph with the lowest point (vertex) at (0, 1). The graph is symmetric around the y-axis, with two linear parts: one with slope 1 for \(x \geq 0\) and one with slope -1 for \(x < 0\), both starting from the vertex.
3Step 3: Evaluating \( \lim_{x \rightarrow -3} g(x) \)
At \( x = -3 \), the function value is \( g(-3) = |-3| + 1 = 3 + 1 = 4 \). Since \( g(x) \) is continuous everywhere, including at \( x = -3 \), the limit \( \lim_{x \rightarrow -3} g(x) = 4 \) as the values of \( g(x) \) approach 4 from both sides as \( x \) approaches -3.
4Step 4: Evaluating \( \lim_{x \rightarrow 0} g(x) \)
As \( x \) approaches 0, both from negative and positive directions, the function approaches \( g(0) = |0| + 1 = 1 \). Since the function is continuous at \( x = 0 \) and has no termination or jump at this point, the limit exists and \( \lim_{x \rightarrow 0} g(x) = 1 \).
Key Concepts
Understanding the Absolute Value FunctionInterpreting the Graph of the Absolute Value FunctionConcept of Continuity in Functions
Understanding the Absolute Value Function
The absolute value function, denoted as \(|x|\), is a special type of function that returns the non-negative value of any given number \(x\). In other words, it measures the "distance" of \(x\) from zero. For example, \(|-3| = 3\) because \(-3\) is three units away from zero.
When you see a function like \(g(x) = |x| + 1\), it means that every output of \(|x|\) is increased by 1. This simply shifts the entire graph of the absolute value function up 1 unit on the y-axis. The V-shape of the basic absolute value graph remains the same but starts from the point (0,1) instead of the origin (0,0).
To visualize this, imagine drawing a V that opens upwards, centered at the origin, and then moving it straight up by one unit without tilting it left or right. This simple transformation highlights the constant positive measure of \(|x|\), no matter if \(x\) is positive or negative.
When you see a function like \(g(x) = |x| + 1\), it means that every output of \(|x|\) is increased by 1. This simply shifts the entire graph of the absolute value function up 1 unit on the y-axis. The V-shape of the basic absolute value graph remains the same but starts from the point (0,1) instead of the origin (0,0).
To visualize this, imagine drawing a V that opens upwards, centered at the origin, and then moving it straight up by one unit without tilting it left or right. This simple transformation highlights the constant positive measure of \(|x|\), no matter if \(x\) is positive or negative.
Interpreting the Graph of the Absolute Value Function
Graph interpretation is a powerful skill in calculus and mathematics in general. When we graph \(g(x) = |x| + 1\), we create a V-shaped graph. This involves two primary lines, showing different behaviors on each side of the y-axis.
The point where these two straight lines meet is the vertex, which in this case is moved up to (0,1) due to the "+1" in the function. This vertex acts as the lowest point of the graph and shows the minimum value of \(g(x)\). Understanding this formation helps us easily apprehend how the graph behaves over the domain of the function.
- For \(x \geq 0\): The graph has a positive slope (slope = 1), indicating that as \(x\) increases, \(g(x)\) increases steadily. This can be seen as a straight line rising to the right from the vertex (0,1).
- For \(x < 0\): The graph has a negative slope (slope = -1), indicating that as \(x\) decreases, \(g(x)\) also decreases, creating a straight line falling to the left from the vertex (0,1).
The point where these two straight lines meet is the vertex, which in this case is moved up to (0,1) due to the "+1" in the function. This vertex acts as the lowest point of the graph and shows the minimum value of \(g(x)\). Understanding this formation helps us easily apprehend how the graph behaves over the domain of the function.
Concept of Continuity in Functions
Continuity is a fundamental concept in calculus. A function is said to be continuous at a point if there are no breaks, jumps, or holes at that point. In simple terms, you can draw the graph of the function without lifting your pencil.
For the function \(g(x) = |x| + 1\), it is continuous everywhere on its domain, which is all real numbers. This is because it is essentially formed by linear parts without any sharp turns or discontinuities.
When evaluating limits like \(\lim_{x \rightarrow -3} g(x)\) and \(\lim_{x \rightarrow 0} g(x)\), we see that both limits exist because as \(x\) approaches these points, \(g(x)\) smoothly approaches a single value — in this case, 4 and 1 respectively.
The continuous nature of \(g(x)\) ensures that these points are solid with no interruptions, allowing us to confidently state the limits. Hence, continuity gives us an assurance of smooth and predictable behavior at every point along the graph.
For the function \(g(x) = |x| + 1\), it is continuous everywhere on its domain, which is all real numbers. This is because it is essentially formed by linear parts without any sharp turns or discontinuities.
When evaluating limits like \(\lim_{x \rightarrow -3} g(x)\) and \(\lim_{x \rightarrow 0} g(x)\), we see that both limits exist because as \(x\) approaches these points, \(g(x)\) smoothly approaches a single value — in this case, 4 and 1 respectively.
The continuous nature of \(g(x)\) ensures that these points are solid with no interruptions, allowing us to confidently state the limits. Hence, continuity gives us an assurance of smooth and predictable behavior at every point along the graph.
Other exercises in this chapter
Problem 54
Find \(y^{\prime}\) $$ \text { If } y=x^{3}+2 x-5, \text { find }\left.\frac{d y}{d x}\right|_{x=-2} $$
View solution Problem 54
Let \(g(x)=\sqrt[3]{x}\). A student graphs this function, and the graph appears to be continuous for all real numbers \(x\). The student concludes that \(g\) is
View solution Problem 54
Tongue-Tied Sauces, Inc., finds that the cost, in dollars, of producing \(x\) bottles of barbecue sauce is given by \(C(x)=375+0.75 x^{3 / 4}\). Find the rate a
View solution Problem 54
Is the function given by $$ F(x)=\left\\{\begin{array}{ll} \frac{1}{3} x+4, & \text { for } x \leq 3 \\ 2 x-5, & \text { for } x>3 \end{array}\right. $$ continu
View solution