Problem 20
Question
Determine whether or not the function is continuous at the given number. $$g(x)=\left\\{\begin{array}{cl} 2 x+5 & \text { if } x<-1 \\ -2 x+1 & \text { if } x \geq-1 \end{array} \text { at } x=-1\right.$$
Step-by-Step Solution
Verified Answer
Answer: Yes, the function g(x) is continuous at x = -1.
1Step 1: Calculate the left-hand limit
To find the left-hand limit, we will calculate the limit as x approaches -1 from the left side (x < -1). We will use the function $$2x + 5$$ as it is defined for x < -1.
The left-hand limit can be written as:
$$\lim_{x \to -1^-} (2x + 5)$$
As we plug in x = -1, we get:
$$\lim_{x \to -1^-} (2x + 5) = 2(-1) + 5 = -2 + 5 = 3$$
2Step 2: Calculate the right-hand limit
To find the right-hand limit, we will calculate the limit as x approaches -1 from the right side (x ≥ -1). We will use the function $$-2x + 1$$ as it is defined for x ≥ -1.
The right-hand limit can be written as:
$$\lim_{x \to -1^+} (-2x + 1)$$
As we plug in x = -1, we get:
$$\lim_{x \to -1^+} (-2x + 1) = -2(-1) + 1 = 2 + 1 = 3$$
3Step 3: Calculate the value of the function at x = -1
The function g(x) is defined at x = -1 using the function $$-2x + 1$$. We can simply plug in x = -1 to find the value of the function at x = -1.
$$g(-1) = -2(-1) + 1 = 2 + 1 = 3$$
4Step 4: Check the continuity conditions
As we have found the left-hand limit, the right-hand limit, and the value of the function at x = -1, we can now check the continuity conditions.
The function g(x) is continuous at x = -1 if the following condition holds:
$$\lim_{x \to -1^-} (2x + 5) = \lim_{x \to -1^+} (-2x + 1) = g(-1)$$
From Step 1, the left-hand limit is 3.
From Step 2, the right-hand limit is 3.
From Step 3, the function value at x = -1 is 3.
All three conditions are equal (3 = 3 = 3), which means the function g(x) is continuous at x = -1.
Key Concepts
Left-hand limitRight-hand limitPiecewise functionLimit evaluationPoint of continuity
Left-hand limit
In mathematics, when we speak about the left-hand limit of a function, we mean the value that the function approaches as the variable gets closer to a specific point from the left side. Imagine you are traveling on a road towards a destination; the left-hand limit is like your approach as you get near the destination from one direction. For the piecewise function given in the exercise, we're interested in how the function behaves as it nears the point where the pieces of the function meet, specifically from values of \( x \) less than -1.
- To calculate the left-hand limit for our piecewise function \( g(x) \), we use the part of the function that is valid for \( x < -1 \), which is \( 2x + 5 \).
- The limit, as \( x \) approaches -1 from the left \((x \to -1^-)\), is given by \( \lim_{x \to -1^-}(2x + 5) \).
- Plugging \(-1\) into the function, \( 2(-1) + 5 = 3 \), allows us to determine that the left-hand limit is \( 3 \).
Right-hand limit
The right-hand limit is another crucial element when analyzing piecewise functions. It represents the value the function approaches as we head towards a specific point from the right.
- For the function \( g(x) \), as \( x \) approaches -1 from the right (values of \( x \) that are −1 or larger), we use the portion defined by \( -2x + 1 \).
- The right-hand limit can be denoted as \( \lim_{x \to -1^+}(-2x + 1) \).
- Substituting \(-1\) into this expression yields \( -2(-1) + 1 = 3 \). Thus, the right-hand limit is \( 3 \).
Piecewise function
Piecewise functions are functions defined by different expressions depending on which interval of the domain the input falls into. They provide flexibility in modeling situations that change at different intervals. A common use is when different scenarios or rules apply in different contexts.
- In our exercise, \( g(x) \) is a piecewise function with two parts, one for \( x < -1 \) and another for \( x \geq -1 \).
- The function rules change at \( x = -1 \), so it's important to understand the behavior from both sides of this breakpoint.
- Piecewise functions require careful evaluation of limits at their points of change to avoid discontinuities at these points.
Limit evaluation
Limit evaluation is the process we use to find the value that a function approaches as the input approaches a particular value. It is an essential concept for understanding properties like continuity.
- Limits help determine how a function behaves near a point, even if the function isn’t explicitly defined at that point.
- They are fundamental in analyzing piecewise functions, providing insights into how different segments of the function meet.
- When evaluating limits for our function \( g(x) \) at \( x = -1 \), we looked at both the left-hand and right-hand limits to ensure they lead to the same result.
- This evaluation confirmed that the function approaches the same value from both sides of the breakpoint.
Point of continuity
A point of continuity for a function is a point where the function is, effectively, seamless. This means that the function doesn't have any jumps, breaks, or holes when moving from one piece to another. In our context, determining if the function \( g(x) \) is continuous at \( x = -1 \) involves checking a few key points.
- Left-hand Limit = Right-hand Limit: Both limits must be equal at the point. For \( g(x) \) at \( x = -1 \), both are \( 3 \).
- Function Value at the Point: The actual function value \( g(-1) \) must also be \( 3 \) for continuity.
- If all these conditions meet, as they do here, the function is continuous at that point of consideration.
Other exercises in this chapter
Problem 19
Find the limit if it exists. If the limit does not exist, explain why. $$\lim _{x \rightarrow 3^{+}} \frac{3}{x^{2}-9}$$
View solution Problem 20
Use the Infinite Limit Theorem and the properties of limits as in Example 6 to find the horizontal asymptotes (if any) of the graph of the given function. $$g(x
View solution Problem 20
Find the limit if it exists. If the limit does not exist, explain why. $$\lim _{x \rightarrow 2^{-}} \frac{x+1}{x^{2}-x-2}$$
View solution Problem 21
Use the Infinite Limit Theorem and the properties of limits to find the limit. $$\lim _{x \rightarrow-\infty} \frac{(x-3)(x+2)}{2 x^{2}+x+1}$$
View solution