Question

Intervals of continuity Complete the following steps for each function. a. Use the continuity checklist to show that \(f\) is not continuous at the given value of \(a\) b. Determine whether \(f\) is continuous from the left or the right at a. c. State the interval(s) of continuity. $$f(x)=\left\\{\begin{array}{ll}2 x & \text { if } x<1 \\\x^{2}+3 x & \text { if } x \geq 1\end{array} ; a=1\right.$$

Short answer

Short answer: No, the function is not continuous at \(x=1\). The intervals of continuity are \((-\infty,1)\cup[1,\infty)\).

Step-by-step solution

Use the continuity checklist to show that \(f\) is not continuous at a=1

A function is continuous at a point \(a\) if it satisfies three conditions: 1. \(f(a)\) is defined 2. \(\lim_{x\to a^-}f(x)\) exists 3. \(\lim_{x\to a^+}f(x)\) exists, and \(\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)\) Function definition: $$ f(x)= \begin{cases} 2x &\text{if}\ x<1 \\ x^{2}+3x &\text{if}\ x\geq 1 \end{cases} $$ Let's use these conditions to determine if \(f\) is continuous at \(a=1\): 1. Evaluate \(f(1)\): $$ f(1)=1^2+3(1)=4 $$ So, the value of the function at \(a=1\) is defined. Now we need to find the left-hand limit and right-hand limit at \(a=1\). 2. Calculate the left-hand limit \(\lim_{x\to1^-}f(x)\): For \(x<1\), \(f(x)=2x\). Thus, we evaluate: $$ \lim_{x\to1^-}2x=2(1)=2 $$ 3. Calculate the right-hand limit \(\lim_{x\to1^+}f(x)\): For \(x\geq1\), \(f(x)=x^2+3x\). Thus, we evaluate: $$ \lim_{x\to1^+}(x^2+3x)=1^2+3(1)=4 $$ Since \(\lim_{x\to1^-}f(x)=2\neq4=\lim_{x\to1^+}f(x)\), the function is not continuous at \(a=1\).

Determine whether \(f\) is continuous from the left or the right at \(a=1\)

In Step 1, we have calculated the left-hand limit and the right-hand limit of the function at \(a=1\): - \(\lim_{x\to1^-}f(x)=2\) - \(\lim_{x\to1^+}f(x)=4\) Since the left-hand limit exists and is different from the right-hand limit, the function is continuous from the left at \(a=1\). However, the function is not continuous from the right at \(a=1\).

State the interval(s) of continuity

Since \(f\) is continuous from the left at \(a=1\), and there are no other discontinuities in the function's definition, we can state the intervals of continuity as follows: $$ (-\infty,1)\cup[1,\infty) $$ In other words, the function is continuous on the entire real line except at the point \(x=1\).

Key concepts

Left-Hand Limit

The concept of a left-hand limit refers to the value that a function approaches as the input approaches a certain point from the left side on the graph. To find the left-hand limit, we observe the values of the function as we get closer and closer to our point of interest, but only from values less than that point. In mathematical terms, we write this as \(\lim_{x\to a^-}f(x)\).
For the exercise given, when we look at the left-hand limit at \(a=1\), we need to consider the part of the function where \(x<1\), which in this case is defined by the expression \(2x\). Therefore, \(\lim_{x\to1^-}f(x)=2\) signifies that as we approach \(x=1\) from the left, the function approaches the value 2. This is crucial for understanding continuity as the behavior of the function from the left directly impacts whether the function is continuous at that point.

Right-Hand Limit

The right-hand limit is similar to the left-hand limit but from the other direction. It considers the behavior of a function as the input comes close to a point from values greater than the point. The right-hand limit is denoted by \(\lim_{x\to a^+}f(x)\). We use it to see what value the function is heading towards as we move towards our point of interest, but now we are coming from the right.
In our exercise, to evaluate the right-hand limit at \(a=1\), we look at the portion of the function where \(x\geq1\) which is given by the formula \(x^2+3x\). Calculating yields \(\lim_{x\to1^+}(x^2+3x)=4\), indicating the value that the function is tending towards is 4 as \(x\) approaches 1 from the right. This information helps us to compare with the left-hand limit to determine overall continuity at that point.

Continuity Checklist

A continuity checklist is a set of criteria used to test whether a function is continuous at a certain point. A function is said to be continuous at a point \(a\) if it meets the following conditions:

• The function \(f(a)\) is defined.
• The left-hand limit \(\lim_{x\to a^-}f(x)\) exists.
• The right-hand limit \(\lim_{x\to a^+}f(x)\) exists.
• Both the left-hand and right-hand limits are equal to each other and to \(f(a)\).

If all these conditions are met, the function is continuous at \(a\). Otherwise, it is said to have a discontinuity there. In the function described in the exercise, we observed that the left-hand and right-hand limits at \(a=1\) do not equal \(f(1)\), which means the function is not continuous at that point, satisfying our investigation using the continuity checklist.

Function Continuity

Function continuity is a property that tells us if a function does not have any abrupt breaks, gaps, or jumps for values within a certain interval. In simple terms, if you can draw the function without lifting your pencil from the paper, then it's continuous across that interval. Continuity of a function is important in calculus because it allows us to use techniques like differentiation and integration which require functions to be continuous in their intervals of interest.
For the given problem, the function \(f\) is continuous for all \(x\) except at \(x=1\). This means there is an interruption in the function's behavior specifically at that point. To express where the function is steady and unbroken, we state its intervals of continuity. As \(f\) is continuous from the left at \(a=1\) and has no other points of disruption, the intervals of continuity are \( (-\infty,1)\cup[1,\infty) \), signifying that only at \(x=1\) the function isn't continuous, aligning with our discovery through the previous steps.