Problem 87 - Calculus: Early Transcendentals

Question

Let $$g(x)=\left\\{\begin{array}{ll}x^{2}+x & \text { if } x<1 \\\a & \text { if } x=1 \\\3 x+5 & \text { if } x>1\end{array}\right.$$ a. Determine the value of \(a\) for which \(g\) is continuous from the left at 1. b. Determine the value of \(a\) for which \(g\) is continuous from the right at 1. c. Is there a value of \(a\) for which \(g\) is continuous at \(1 ?\) Explain.

Step-by-Step Solution

Answer

Answer: No, there is no value of \(a\) that makes the function \(g(x)\) continuous at \(x=1\).

Step 1: Find the limit of the function from the left-hand side of 1.

To find the limit from the left, we will approach 1 from the left using the function definition \(x^2+x\). $$\lim_{x\to1^-}g(x)=\lim_{x\to1^-}(x^2+x)$$

Step 2: Evaluate the limit from the left-hand side of 1.

To evaluate the limit, we substitute 1 into the function definition: $$\lim_{x\to1^-}g(x)=\lim_{x\to1^-}(1^2+1)=1+1=2$$ So, the limit from the left-hand side of 1 is 2.

Step 3: Equate the limit to the function value at 1.

The function is continuous from the left if the limit from the left-hand side equals the function value at \(x=1\). $$a=\lim_{x\to1^-}g(x)=2$$ Thus, the value of \(a=2\) for which \(g\) is continuous from the left at 1. #b. Determine the value of \(a\) for which \(g\) is continuous from the right at 1.#

Step 1: Find the limit of the function from the right-hand side of 1.

To find the limit from the right, we will approach 1 from the right using the function definition \(3x+5\). $$\lim_{x\to1^+}g(x)=\lim_{x\to1^+}(3x+5)$$

Step 2: Evaluate the limit from the right-hand side of 1.

To evaluate the limit, we substitute 1 into the function definition: $$\lim_{x\to1^+}g(x)=\lim_{x\to1^+}(3(1)+5)=3+5=8$$ So, the limit from the right-hand side of 1 is 8.

Step 3: Equate the limit to the function value at 1.

The function is continuous from the right if the limit from the right-hand side equals the function value at \(x=1\). $$a=\lim_{x\to1^+}g(x)=8$$ Thus, the value of \(a=8\) for which \(g\) is continuous from the right at 1. #c. Is there a value of \(a\) for which \(g\) is continuous at \(1?\) Explain.#

Step 1: Check continuity at \(x=1\)

A function is continuous at a point if there exists a limit at that point, and the limit equals the function value at that point. We already found that for left continuity \(a=2\), and for right continuity \(a=8\). Since the value of \(a\) cannot be equal to both 2 and 8 at the same time, we conclude that there is no value of \(a\) for which the function is continuous at \(x=1\).

Key Concepts

Left-Hand Limit

The concept of the left-hand limit comes into play when exploring what happens to a function as we approach a specific point from values smaller than that point. It's denoted by the limit notation \( \lim_{x \to c^-} f(x) \), where \( c \) is the point of interest. In plain terms, it asks, "What value does the function get close to, as we inch towards \( c \) from values less than \( c \)?"

For example, in our function \( g(x) \), to find the left-hand limit as \( x \) approaches 1, we look at the piece of the piecewise function defined by \( x^2 + x \) for \( x<1 \). By plugging in values of \( x \) that get closer to 1 from the left, like 0.9 or 0.99, and evaluating \( x^2+x \), we see the result gets closer to 2. Mathematically, the limit is found to be 2 when calculated directly by substituting \( x=1 \) as shown:

• \( \lim_{x \to 1^-} (x^2 + x) = 2 \)

For this reason, when discussing the continuity of piecewise functions from the left, the function value needs to equal this left-hand limit to ensure smooth, unbroken movement across the transition.

Right-Hand Limit

The right-hand limit is similar to the left-hand limit, but focuses on the behavior of a function as you approach a specific point from values larger than that point. This is expressed with the limit notation \( \lim_{x \to c^+} f(x) \), where \( c \) is the point you are approaching. Simply put, it inquires, "What value does the function tend to as we come close to \( c \) from the right?"

For piecewise functions such as \( g(x) \), when determining the right-hand limit as \( x \) nears 1, we use the definition \(3x+5\) for \( x > 1 \). By evaluating \( 3x + 5 \) at values just greater than 1 like 1.1 or 1.01, the function approaches the result 8. Plugging in 1 directly yields:

• \( \lim_{x \to 1^+} (3x + 5) = 8 \)

It's essential for the function to match this right-hand limit at the point to ensure right continuous behavior. When this condition isn't met, it indicates a break or jump, resulting in a sudden change in function value as you pass through \( x=1 \) from the right.

Piecewise Function

A piecewise function is a type of function characterized by different expressions depending on the input value. This means that the function 'pieces' together different equations to cover various intervals of the domain of the function. They are ideal for modeling real-world scenarios where a rule or relationship changes at certain points within their domain.

In the example of \( g(x) \), the piecewise definition is:

• \( x^2 + x \) for \( x < 1 \)
• \( a \) for \( x = 1 \)
• \( 3x + 5 \) for \( x > 1 \)

This formulation allows each segment to dictate the function’s value over specified ranges. For piecewise functions to be called continuous, both the left-hand and right-hand limits at the boundary, and the actual value of the function at the boundary, must coincide. In other words, at \( x=1 \), the limits and function value \( a \) need to match so there's no jump or gap.

Understanding how these pieces interact is crucial, making it clear why specific continuity conditions like equal left-hand and right-hand limits, as well as a matching function value, are necessary.