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.

Short answer

Explain. Answer: No, there is no value of 'a' that will make the function 'g' continuous at x = 1. This is because the left limit at x = 1 equals 2, and the right limit equals 8. Since the left and right limits are different, it is not possible to find a single value of 'a' that satisfies the condition of continuity at x = 1.

Step-by-step solution

Identify the left and right expressions of the function

The function is defined by three expressions. We are given the task of making 'g' continuous from the left at x = 1: - \(g(x) = x^2 + x\) for \(x < 1\) - \(g(x) = a\) for \(x = 1\) - \(g(x) = 3x + 5\) for \(x > 1\) Since we're looking for continuity from the left, we need to focus on the interval where x<1, which has the expression \(x^2 + x\).

Determine the limit from the left

We need to find the limit from the left at x=1, which is expressed as \(\lim_{x \to 1^-}(x^2+x)\). Using direct substitution, we get: \(\lim_{x \to 1^-}(x^2+x) = (1)^2 + 1 = 2\)

Make the function continuous from the left at x = 1

In order for the function to be continuous from the left at x = 1, we need the left limit and the value of the function at x = 1 to be equal. Using the left limit we calculated in step 2 and knowing that \(g(1) = a\), we get: \(2 = a\) So, the value of \(a\) for which \(g\) is continuous from the left at x = 1 is 2. #b. Determine the value of \(a\) for which \(g\) is continuous from the right at 1.#

Determine the limit from the right

In a similar manner, we now need to find the limit from the right at x=1, which is expressed as \(\lim_{x \to 1^+}(3x+5)\). Using direct substitution, we get: \(\lim_{x \to 1^+}(3x+5) = 3(1) + 5 = 8\)

Make the function continuous from the right at x = 1

We need the right limit, and the value of the function at x = 1 to be equal for the continuity from the right: \(8 = a\) So, the value of \(a\) for which \(g\) is continuous from the right at x = 1 is 8. #c. Is there a value of \(a\) for which \(g\) is continuous at 1? Explain.#

Test for continuity at x = 1

In order for \(g\) to be continuous at x = 1, the left and right limits must be equal, and they both must equal the value of the function at x = 1. However, as we saw in steps 3 and 5, the left limit equals 2 and the right limit equals 8: \(a = 2\) for continuity from the left \(a = 8\) for continuity from the right Since the left and right limits are different, there is no value of 'a' that will make the function continuous at x = 1.

Key concepts

Left-hand Limit

In calculus, the left-hand limit of a function at a certain point is the value the function approaches as you get closer to that point from the left side. Imagine walking towards a cliff's edge from the left side—where would you be just before stepping off? That's the idea behind the left-hand limit.

Mathematically, if you have a function like in our example, where it's defined differently based on the value of x, you are looking for the limit of the function as x approaches a certain value, say 1, from numbers smaller than 1. In other words, for any small positive value \( \epsilon \), the function's value at \( 1 - \epsilon \) is the left-hand limit as \( \epsilon \) gets closer and closer to 0. It's represented as \( \lim_{{x \to 1^-}} g(x) \).

The concept of left-hand limit is essential in understanding the continuity of functions at a point. A function is continuous at a point only if the left-hand limit equals the function's value at that point, which also must equal the right-hand limit—a beautiful harmony in mathematics.

Right-hand Limit

Conversely, the right-hand limit is what the function approaches as you get closer to a point from the right side. Think about that cliff again—now approaching from the right. The right-hand limit tells us the value right before the 'step off’ from the right.

For a function defined differently on each side of a point, we check the part of the function that applies when x is slightly more than our point of interest. Mathematically, it's the function's value at \( 1 + \epsilon \) for a tiny positive \( \epsilon \), and we denote this limit as \( \lim_{{x \to 1^+}} g(x) \).

Understanding the right-hand limit is crucial for determining if a function is continuous from the right at a specific point. It plays a role analogous to the left-hand limit, and for perfect continuity at a point, both must match and also equal the function's value at that point.

Piecewise Functions

Let's dive into the world of piecewise functions. These are like a patchwork quilt, made of different fabric patches—each piece of the function is defined by a distinct rule over different intervals. Just as each patch in a quilt has a pattern that fits into the overall design, each piece of a piecewise function has a formula that applies to certain values of x.

They can be tricky because the overall behavior of the function depends on which piece you're looking at. In the case of our example, we have a function \( g(x) \) that has three pieces, behaving differently for x values less than 1, exactly at 1, and greater than 1. For these types of functions, examining limits and continuity requires careful attention to these boundaries—where the pieces 'meet'. An exciting challenge is finding the right 'patch' that will make the whole 'quilt' of the function continuous.

Limits and Continuity

The concept of limits and continuity is a cornerstone of calculus. Limits are about approaching a point, whereas continuity is about being smooth and unbroken at that point. To visualize this, picture a roller coaster. The limit is the height you nearly reach with your hands up as you approach the top of a hill, and continuity ensures you can actually ride over the hill without any sudden drops or gaps.

In the practice of calculus, we need a function's left-hand and right-hand limits to agree and match the function's actual value at the point to claim it's continuous there. This agreement is what makes functions predictable and nice to work with. In our textbook problem, the search for continuity at x=1 involves finding a value of 'a' that seamlessly stitches the left and right behaviors of \( g(x) \), which, as we found out, isn't possible since limits from the left and the right do not match.

Continuity maintains the flow of a function, without any jumps or holes, ensuring a smooth ride through the mathematical journey.