Lesson Explainer: Continuity of Functions | Nagwa Lesson Explainer: Continuity of Functions | Nagwa

Lesson Explainer: Continuity of Functions Mathematics

In this explainer, we will learn how to check the continuity of a function over its domain and determine the interval on which it is continuous.

Let us begin by recalling what it means for a function to be continuous at a point.

Definition: Continuity of a Function at a Point

Let 𝑎. We say that a real-valued function 𝑓(𝑥) is continuous at 𝑥=𝑎 if lim𝑓(𝑥)=𝑓(𝑎).

This definition requires the function to satisfy the following three conditions:

  1. 𝑓(𝑎) is defined;
  2. lim𝑓(𝑥) exists;
  3. 𝑓(𝑎)=𝑓(𝑥)lim.

Each of these conditions listed above has geometric implications in the graphs of the function. Understanding these implications will help us understand what it means for a function to be continuous on an interval.

The first condition of continuity at 𝑥=𝑎 states that 𝑓(𝑎) should be defined. This condition can be seen as a no-hole condition. We can understand this condition better by considering the opposite scenario, where 𝑓(𝑎) is not defined. This means that 𝑎 is not in the domain of the function, which means that the graph of this function will not intersect with the vertical line 𝑥=𝑎. Generically, we can think of these graphs as having an open circle at 𝑥=𝑎.

For instance, consider a function 𝑓(𝑥)=𝑥+1 with the domain {1}. The graph of 𝑦=𝑥+1 is a straight line, but we can see that the domain of this function excludes 𝑥=1. This fact can be indicated by placing an open circle on the straight line at 𝑥=1, as seen below.

If we consider the same function with the domain , the graph will not have an open circle. Hence, this condition is often referred to as the no-hole condition.

The second condition of continuity states that the limit of 𝑓(𝑥) at 𝑥=𝑎 exists, which can be considered as a no-gap (and no-oscillation) condition. Recall that a limit of a function does not exist if

  • the limit at the point is unbounded,
  • the one-sided limits at the point are not equal, or
  • the function oscillates at the point.

The last case is a special case that does not appear often in examples, and we can see this behavior in the following graph of a function.

We can see that as 𝑥 approaches 0, the 𝑦-coordinates of points on the graph oscillate between 1 and 1 more and more rapidly. Since this value does not approach a specific value, the limit of this function as 𝑥 approaches 0 does not exist.

Besides the last scenario, the first two cases mean that there will be a gap between the graph of the function on the left and the right side of 𝑥=𝑎. For instance, we can consider the function 𝑔(𝑥)=1𝑥<0,1𝑥0.ifif

We know that the limit of this function at 𝑥=0 does not exist since the one-sided limits are not equal to each other. As a consequence of this fact, there is a gap or a jump in the graph of the function at 𝑥=0. If the limit of a function exists at a point, the graph will not have a gap (or an oscillation) at the point. Hence, this condition is referred to as the no-gap (no-oscillation) condition.

The third condition of continuity states that the limit of the function at 𝑥=𝑎 must equal the function value at 𝑥=𝑎. This condition also corresponds to the no-hole condition. We recall that the limit of a function at 𝑥=𝑎 describes the behavior of the function near 𝑥=𝑎, but not at 𝑎. Hence, this condition tells us that, as we get closer to 𝑥=𝑎, the function value will approach the 𝑦-coordinate of the point on the graph at 𝑥=𝑎. Let us consider the opposite scenario, when this condition is not met. Consider the graph of the function (𝑥)=𝑥+1𝑥1,1𝑥=1.ifif

From the given graph, we can see that the limit of (𝑥) at 𝑥=1 is 2. On the other hand, the value of (1) is given by the 𝑦-coordinate of the solid point, which is equal to 1. Since 21, the limit of 𝑓(𝑥) at 𝑥=1 does not equal (1) for this function. As a result, we see that there is a hole, or an open circle, in the graph of (𝑥) at 𝑥=1. The difference compared to the graph of 𝑓(𝑥) above is that the graph of (𝑥) contains a closed circle at (1,1) in the graph. Both graphs have an open circle, or a hole, at (1,2).

We can think of the case where all three conditions are met, which would lead to a function that is continuous at 𝑥=𝑎. Consider the graph of the function 𝑗(𝑥)=𝑥+1𝑥1,2𝑥=1.ifif

At 𝑥=1, we can see that 𝑗(1) is defined, and the limit of this function at 𝑥=1 exists. We can see from the graph that lim𝑗(𝑥)=2,𝑗(1)=2.

This tells us that the third condition for continuity at 𝑥=1 is satisfied. Hence, this function is continuous at 𝑥=1. As a result, there is no hole and no gap (and no oscillation) in the graph of this function at 𝑥=1.

We now define the continuity of a function on an interval.

Definition: Continuity of a Function on an Interval or a Set

We say that a function is continuous on an interval (or a set) if it is continuous at each point in the interval (or the set).

What does the continuity on an interval say about the graph of the function? If a function is continuous on an interval, it is continuous at each point in the interval. This tells us that the graph of this function cannot have any holes or gaps over this interval. In other words, the graph of this function is a connected curve. Loosely speaking, this means that we can draw the graph of a continuous function without lifting our pencil from the paper.

In our first example, we will determine whether a function is continuous over an interval using its graph.

Example 1: Identifying Continuous and Discontinuous Functions Graphically

Determine whether the function represented by the graph is continuous or discontinuous on the interval [0,3].

Answer

In this example, we need to determine whether a function is continuous over an interval by the given graph. We will demonstrate how to determine the continuity of a function, first, using heuristics and, second, definitions.

Method 1

We know that a function is continuous on an interval if the graph of the function does not have any holes or gaps over the interval. Loosely speaking, this means that we can draw the graph of a continuous function without lifting our pencil from the paper. We can draw the given graph without lifting our pencil from a paper because the graph is connected.

This tells us that the function represented by the graph is continuous.

Method 2

We recall that a function is continuous on an interval if it is continuous at every point in the interval. Hence, we need to determine whether 𝑓(𝑥) is continuous at 𝑥=𝑎 for each 𝑎[0,3]. We recall that 𝑓(𝑥) is continuous at 𝑥=𝑎 if it satisfies the following three conditions:

  1. 𝑓(𝑎) is defined;
  2. lim𝑓(𝑥) exists;
  3. 𝑓(𝑎)=𝑓(𝑥)lim.

We can see that 𝑓(𝑎) is defined for each number 𝑎 in the interval [0,3] since we can find a point on the graph of the function for each 𝑥-coordinate 𝑎. The 𝑦-coordinate of this point gives 𝑓(𝑎). This means that the first condition is met.

For the second condition, we can see that the limit of the function exists at each 𝑥-value within this interval since following the graph of the function as 𝑥 approaches 𝑎 always results in a specific point. The 𝑦-coordinate of the point that we reach is the limit of 𝑓(𝑥) at 𝑥=𝑎. Thus, the second condition is met.

We can see that the point that we reach when finding the limit of 𝑓(𝑥) at 𝑥=𝑎 lies on the graph of the function, which means that the limit of the function and the value of the function at 𝑥=𝑎 are both equal to the 𝑦-coordinate of this point. Hence, these two values must be equal, meaning that the third condition is met.

Since all three conditions of continuity at a point are met for each number 𝑎 in the interval [0,3], we know that 𝑓(𝑥) is continuous at each point in the interval.

Hence, the function represented by the graph is continuous on the interval [0,3].

In the previous example, we determined whether a function was continuous on the interval by examining its graph. We say that a function is continuous everywhere if it is continuous on , or equivalently ],[. In general, when we refer to a function as being continuous without reference to a specific point or interval, we mean that the function is continuous everywhere.

Let us consider another example of determining whether a function is continuous from its graph.

Example 2: Determining Whether a Given Graph Is Continuous

True or False: The function in the given graph is a continuous function.

Answer

In this example, we need to determine whether a function is continuous by the given graph. Recall that, when we refer to a function as being continuous without reference to a specific point or interval, we mean that the function is continuous everywhere. A function is continuous everywhere if it is continuous at every point.

We will demonstrate how to determine the continuity of a function, first, using heuristics and, second, definitions.

Method 1

We know that a function is continuous on an interval if the graph of the function does not have any holes or gaps over the interval. Loosely speaking, this means that we can draw the graph of a continuous function without lifting our pencil from the paper. We cannot draw the given graph without lifting our pencil from a paper because the graph contains a gap at 𝑥=3. We will have to lift our pencil when we move from the left side of 𝑥=3 to the right side of 𝑥=3. This tells us that the function in the given graph is not continuous.

Hence, the given statement is false.

Method 2

We recall that a function is continuous everywhere if it is continuous at every point. Hence, we need to determine whether 𝑔(𝑥) is continuous at 𝑥=𝑎 for each 𝑎. We recall that 𝑔(𝑥) is continuous at 𝑥=𝑎 if it satisfies the following three conditions:

  1. 𝑔(𝑎) is defined;
  2. lim𝑔(𝑥) exists;
  3. 𝑔(𝑎)=𝑔(𝑥)lim.

We can see that 𝑔(𝑎) is defined for each real number 𝑎 since we can find a point on the graph of the function for each 𝑥-coordinate 𝑎. The 𝑦-coordinate of this point gives 𝑔(𝑎). This means that the first condition is met.

Next, let us consider the second condition. Recall that the (regular) limit of a function does not exist if the one-sided limits do not equal each other. From the given graph, we can see that the left-hand limit of 𝑔(𝑥) at 𝑥=3 is equal to 2, while the right-hand limit at this point is equal to 1. This means that limlim𝑔(𝑥)=2,𝑔(𝑥)=1.

Since 21, the left and right limits do not agree. This tells us that the limit of this function does not exist at 𝑥=3 so the second condition of continuity is not met at 𝑥=3. Hence, the function in the given graph is not continuous at 𝑥=3.

The given statement is false.

When we consider the graph of a function, we can determine its continuity as we have seen in previous examples. If we know the graph of the function, we can determine whether or not it is continuous on an interval by asking if it is possible to draw the graph without lifting a pencil from the paper. Of course, this is an informal process and it does not work in every situation since there are graphs of continuous functions that are impossible to draw by hand. However, these are exceptions that we do not need to dwell on at this time. This informal process works very well in generic cases.

For instance, we know that the graph of any polynomial is a connected curve and its domain contains all real numbers, which means that a polynomial function is continuous everywhere. Polynomials are not the only functions with this property, and we can summarize the list of familiar functions that are continuous everywhere.

Property: Functions That Are Continuous Everywhere

The following types of functions are continuous everywhere:

  • a constant function,
  • a polynomial function,
  • an exponential function,
  • the absolute value function,
  • an odd root function (e.g., cube root),
  • the sine and cosine functions.

Let us consider a few examples where we will determine whether a given function is continuous by identifying its type.

Example 3: Discussing the Continuity of a Polynomial Function

What can be said about the continuity of the function 𝑓(𝑥)=𝑥+5𝑥2𝑥+2?

  1. 𝑓(𝑥) is discontinuous on because it is a polynomial.
  2. 𝑓(𝑥) is continuous on because it is a polynomial.

Answer

We can see that 𝑓(𝑥) is a polynomial function. Recall that a polynomial function is continuous on ; hence, 𝑓(𝑥) must be continuous on .

Another way to reach this conclusion is by recalling that a function is continuous if we can draw its graph on a paper without lifting our pencil from the paper. Since the graph of a polynomial is a connected curve, we can draw its graph without lifting our pencil from the paper. This tells us that 𝑓(𝑥) is continuous on because it is a polynomial.

This is option B.

Previously, we identified a list of functions that are continuous everywhere. But there are a lot of functions that we know that are not mentioned in this list. For instance, we are familiar with rational functions, which are not listed as a continuous function. We know that the graph of a rational function may not consist of a single connected curve, especially if the function has vertical asymptotes. These vertical asymptotes occur outside the domain of rational functions. If we restrict our attention to the domain of rational functions, we can see that their graphs are always connected within its domain.

In fact, this is true for almost all of the functions that we are familiar with, except for piecewise-defined functions. In most generic functions, we can draw their graphs within each connected interval of their domains without lifting a pencil from the paper. This leads to the following list of functions that are continuous in their domains.

Property: Functions That Are Continuous on their Domains

The following types of functions are continuous on their domains:

  • a rational function,
  • a logarithmic function,
  • an even root function (e.g., square root),
  • the tangent function.

If a number is not in the domain of a function, we know that the function is not continuous there because the first condition of continuity is not met at that point. Hence, for the functions listed above, finding the set on which they are continuous is equivalent to finding the domains of the functions.

In the next example, we will find the set on which a given rational function is continuous.

Example 4: Discussing the Continuity of Rational Functions

Find the set on which 𝑓(𝑥)=𝑥22𝑥2𝑥63 is continuous.

  1. 𝑓(𝑥) is continuous on {22}.
  2. 𝑓(𝑥) is continuous on {9,7}.
  3. 𝑓(𝑥) is continuous on .
  4. 𝑓(𝑥) is continuous on {9,7}.

Answer

In this example, we need to determine the set on which a rational function is continuous. Recall that a rational function is continuous on its domain. So the set on which 𝑓(𝑥) is continuous is the same as its domain. Let us find the domain of 𝑓(𝑥).

Recall that the domain of a rational function is the set of all real numbers except for the roots of the denominator. In the given function, the denominator is the quadratic polynomial 𝑥2𝑥63. We need to find the roots of this quadratic function.

We can find the root of the quadratic function by first factoring 𝑥2𝑥63=(𝑥9)(𝑥+7).

This tells us that the root of the denominator of the given rational function is 𝑥=9 and 7. Hence, the domain of 𝑓(𝑥) is {9,7}, which is also the set on which 𝑓(𝑥) is continuous.

This is option B.

We recall the property of continuity of a function at a point.

Property: Continuity of Functions at a Point

Let 𝑓(𝑥) and 𝑔(𝑥) be continuous at 𝑥=𝑎. Then,

  • functions 𝑓(𝑥)+𝑔(𝑥), 𝑓(𝑥)𝑔(𝑥), and 𝑓(𝑥)𝑔(𝑥) are continuous at 𝑥=𝑎,
  • 𝑓(𝑥)𝑔(𝑥) is continuous at 𝑥=𝑎 if 𝑔(𝑎)0.

Also, if 𝑓(𝑥) and 𝑔(𝑦) are continuous at 𝑥=𝑎 and 𝑦=𝑓(𝑎), respectively. Then the composite function 𝑔(𝑓(𝑥)) is continuous at 𝑥=𝑎.

Since the continuity of a function on an interval is defined as the continuity of the function at each point in the interval, we can see that similar properties will hold for continuity on a interval, as we can see in the following.

Property: Continuity for the Sum, Difference, Product, Quotient, and Composition of Functions

Let 𝑓(𝑥) and 𝑔(𝑥) be continuous on sets 𝑆 and 𝑇 respectively. Then,

  • functions 𝑓(𝑥)+𝑔(𝑥), 𝑓(𝑥)𝑔(𝑥), and 𝑓(𝑥)𝑔(𝑥) are continuous on 𝑆𝑇,
  • 𝑓(𝑥)𝑔(𝑥) is continuous on the set 𝑆𝑇𝐴, where 𝐴 is the set 𝐴={𝑥𝑔(𝑥)=0},
  • 𝑓𝑔(𝑥) is continuous on the set 𝐵𝑇 where 𝐵 is the set 𝐵={𝑥𝑔(𝑥)𝑆}.

In the next example, we will determine the set on which a sum of two rational functions is continuous.

Example 5: Discussing the Continuity of the Sum of Two Rational Functions

Find the set on which 𝑓(𝑥)=4𝑥+10𝑥+9 is continuous.

  1. The function 𝑓 is continuous on .
  2. The function 𝑓 is continuous on {0,3}.
  3. The function 𝑓 is continuous on {3}.
  4. The function 𝑓 is continuous on {0}.
  5. The function 𝑓 is continuous on {0,3}.

Answer

In this example, we need to determine the set on which the sum of two rational functions is continuous. The first function does not appear to be a rational function, but we can rewrite this function as 4𝑥. Thus, the given function is the sum of 𝑔(𝑥)=4𝑥 and (𝑥)=10𝑥+9.

Recall that the sum of two functions is continuous on the intersection of the sets on which either function is continuous. We can begin by identifying the sets on which 𝑔(𝑥) and (𝑥) are continuous.

Let us begin with 𝑔(𝑥). We know that a rational function is continuous on its domain. So the set on which 𝑔(𝑥) is continuous is the same as its domain. The domain of a rational function is the set of all real numbers except for the roots of the denominator. In 𝑔(𝑥), the denominator is 𝑥, which has one root 𝑥=0. This leads to the domain of 𝑔(𝑥): {0}.

We note that this is also the set on which 𝑔(𝑥) is continuous.

Let us consider (𝑥). The denominator of (𝑥) is 𝑥+9. To find the root of the denominator, we need to solve the equation 𝑥+9=0.

Rearranging this equation leads to 𝑥=9. But we know that a square of a real number cannot be negative, which tells us that this equation does not have a real solution. This tells us that the domain of (𝑥) is , which is also the set on which (𝑥) is continuous.

For the sum of these functions to be continuous, we need to find the intersection of these two sets: ({0})={0}.

This is option D.

Let us turn our focus to piecewise-defined functions. A piecewise-defined function usually consists of subfunctions defined in disjoint intervals. Subfunctions are usually familiar functions such as those in the list of continuous functions (whether everywhere or within their respective domains). This means that, generally, we only need to examine closely the continuity of the piecewise-defined function on the endpoints (or the boundaries) of the disjoint intervals.

In the next example, we will consider where a given piecewise-defined function is continuous.

Example 6: Discussing the Continuity of a Piecewise-Defined Function Involving Trigonometric Ratios at a Point

Discuss the continuity of the function 𝑓, given 𝑓(𝑥)=3+𝑥,0𝑥<𝜋2,4+𝑥𝜋2,𝑥𝜋2.sin

Answer

In this example, we need to discuss the continuity of a piecewise-defined function. We know that a function is never continuous outside its domain, so let us begin by finding the domain of this function. For 𝑓(𝑥) to be defined, 𝑥 needs to satisfy either condition 0𝑥<𝜋2𝑥𝜋2.or

This means that the union of these intervals, which is [0,[, is the domain of 𝑓(𝑥). Now, we need to determine whether 𝑓(𝑥) is continuous at each point in the domain.

The given piecewise-defined function has two subfunctions: 𝑔(𝑥)=3+𝑥0,𝜋2,(𝑥)=4+𝑥𝜋2𝜋2,.sinontheintervalontheinterval

We can see that 𝑔(𝑥) is the sum of a constant and the sine function. We know that the constant function and the sine function are continuous everywhere and the sum of two continuous functions is continuous. Hence, 𝑔(𝑥) is continuous. (𝑥) is a polynomial, and we recall that a polynomial function is continuous.

Since both subfunctions are continuous functions, we know that the given function 𝑓(𝑥) is continuous away from the endpoints of the disjoint intervals 0,𝜋2 and 𝜋2,. Hence, we only need to consider the continuity of 𝑓(𝑥) at the endpoints of these intervals, which are 𝑥=0 and 𝑥=𝜋2.

Let us consider first the point 𝑥=0. We can see that 𝑓(𝑥) is not defined on the left side of this point, 𝑥<0. Hence, the continuity of 𝑓(𝑥) at 𝑥=0 only depends on the continuity of the subfunction 𝑔(𝑥) at 𝑥=0. Since 𝑔(𝑥) is continuous everywhere, 𝑓(𝑥) is continuous at 𝑥=0.

Next, we consider the point 𝑥=𝜋2. To determine the continuity of 𝑓(𝑥) at this point, we recall the three conditions for continuity: 𝑓(𝑥) is continuous at 𝑥=𝑎 if it satisfies the following three conditions:

  1. 𝑓(𝑎) is defined;
  2. lim𝑓(𝑥) exists;
  3. 𝑓(𝑎)=𝑓(𝑥)lim.

Since 𝑥=𝜋2 satisfies the second piecewise condition 𝑥𝜋2, we know that 𝑓𝜋2 is defined using the second subfunction (𝑥). In particular, we can compute 𝑓𝜋2=𝜋2=4+𝜋2𝜋2=4.

Hence, the first condition of continuity is met. To consider the second condition, we recall that the (regular) limit of a function at a point exists if and only if the one-sided limits are equal to each other. Let us determine the one-sided limits of 𝑓(𝑥) at 𝑥=𝜋2.

The left-handed limit considers the values of 𝑓(𝑥) on the left side of 𝑥=𝜋2, which means 𝑥<𝜋2. These values of 𝑥 satisfy the first piecewise condition, so we can compute 𝑓(𝑥) by using the first subfunction 𝑔(𝑥). Hence, limlimlimsin𝑓(𝑥)=𝑔(𝑥)=(3+𝑥).

This is a one-sided limit of the sum of a constant function and the sine function. Recall that we can find the limit (regular or one-sided) of such functions by direct substitution. This leads to limsinsin(3+𝑥)=3+𝜋2=3+1=4.

This gives us lim𝑓(𝑥)=4.

Next, we need to find the right-handed limit. The right-handed limits consider the values of 𝑓(𝑥) on the right side of 𝑥=𝜋2, which means 𝑥>𝜋2. For these values of 𝑥, 𝑓(𝑥) is the same as (𝑥), which leads to limlimlim𝑓(𝑥)=(𝑥)=4+𝑥𝜋2.

Recall that we can find the limit (regular or one-sided) of a polynomial by direct substitution. Hence, lim4+𝑥𝜋2=4+𝜋2𝜋2=4.

This gives us lim𝑓(𝑥)=4.

Since the left and the right limits of 𝑓(𝑥) at 𝑥=𝜋2 are the same, we know that the (regular) limit exists and is equal to this value. This leads to lim𝑓(𝑥)=4.

This tells us that the second condition of continuity is met.

We have already computed 𝑓𝜋2=4,𝑓(𝑥)=4.lim

This leads to the third condition 𝑓𝜋2=𝑓(𝑥).lim

Since all three conditions of continuity are met, 𝑓(𝑥) is continuous at 𝑥=𝜋2.

Finally, we have observed that 𝑓(𝑥) is continuous at each number in its domain [0,[. Thus, the function 𝑓 is continuous on [0,[.

In our final example, we will identify an unknown constant in a piecewise-defined function by using the fact that the function is continuous.

Example 7: Finding the Unknown That Makes a Piecewise-Defined Function Continuous over Its Domain

Find the value 𝑎 that makes the function 𝑓 continuous on its domain given 𝑓(𝑥)=𝑎7𝑥,𝑥2,𝑎𝑥+7,𝑥>2.

Answer

In this example, we need to identify the value of 𝑎 that makes the function 𝑓 continuous on its domain. Let us begin by finding the domain of this function. For 𝑓(𝑥) to be defined, 𝑥 needs to satisfy either condition 𝑥2𝑥>2.or

This means that the union of these intervals, which is , is the domain of 𝑓(𝑥). Now, we need to choose 𝑎 so that 𝑓(𝑥) is continuous at any number.

𝑓(𝑥) has two subfunctions: 𝑔(𝑥)=𝑎7𝑥],2],(𝑥)=𝑎𝑥+7]2,[.ontheintervalontheinterval

Both 𝑔(𝑥) and (𝑥) are polynomials, and we can recall that a polynomial function is continuous everywhere. This means that both 𝑔(𝑥) and (𝑥) are continuous everywhere. Since both subfunctions are continuous functions, we know that the given function 𝑓(𝑥) is continuous away from the endpoint of the disjoint intervals ],2] and ]2,[, which is the point 𝑥=2. In other words, we know that 𝑓(𝑥) is continuous on {2} regardless of the value of 𝑎. Hence, the choice of 𝑎 should be such that 𝑓(𝑥) is continuous at 𝑥=2.

Let us consider the continuity of 𝑓(𝑥) at 𝑥=2. Recall the three conditions for 𝑓(𝑥) being continuous at 𝑥=𝑎:

  1. 𝑓(𝑎) is defined;
  2. lim𝑓(𝑥) exists;
  3. 𝑓(𝑎)=𝑓(𝑥)lim.

Since 𝑥=2 satisfies the first piecewise condition 𝑥2, we know that 𝑓(2) is defined using the first subfunction 𝑔(𝑥). Therefore, we can compute 𝑓(2)=𝑔(2)=𝑎7×2=𝑎14.

Hence, the first condition of continuity is met, regardless of the choice of 𝑎. To consider the second condition, we recall that the (regular) limit of a function at a point exists if and only if the one-sided limits are equal to each other. Let us determine the one-sided limits of 𝑓(𝑥) at 𝑥=2.

The left-handed limits consider the values of 𝑓(𝑥) on the left side of 𝑥=2, which means 𝑥<2. These values of 𝑥 satisfy the first piecewise condition, so we can compute 𝑓(𝑥) by using the first subfunction 𝑔(𝑥). Hence, limlimlim𝑓(𝑥)=𝑔(𝑥)=(𝑎7𝑥).

This is a one-sided limit of a polynomial function. Recall that we can find the limit (regular or one-sided) of a polynomial by direct substitution. This leads to lim(𝑎7𝑥)=𝑎7×2=𝑎14.

This gives us lim𝑓(𝑥)=𝑎14.

Next, we need to find the right-handed limit. The right-handed limits consider the values of 𝑓(𝑥) on the right side of 𝑥=2, which means 𝑥>2. For these values of 𝑥, 𝑓(𝑥) is the same as (𝑥), which leads to limlimlim𝑓(𝑥)=(𝑥)=𝑎𝑥+7.

Using direct substitution, we can write lim𝑎𝑥+7=𝑎×2+7=4𝑎+7.

This gives us lim𝑓(𝑥)=4𝑎+7.

In order to satisfy the second condition of continuity, we need to make sure that the left and the right limits are equal. This means 𝑎14=4𝑎+7.

Rearranging this equation leads to 3𝑎=21, which gives 𝑎=7. This tells us that the second condition will only be met if 𝑎=7. With this choice of 𝑎, the regular limit exists and is equal to the one-sided limit. Hence, lim𝑓(𝑥)=4×(7)+7=21.

Also, 𝑓(2)=714=21.

This tells us that, with 𝑎=7, we have 𝑓(2)=𝑓(𝑥).lim

Since all three conditions of continuity are met with 𝑎=7, 𝑓(𝑥) is continuous at 𝑥=2 with this choice of 𝑎. Together with our previous conclusion that 𝑓(𝑥) is continuous on {2} regardless of the choice of 𝑎, we can conclude that 𝑓(𝑥) is continuous on if we choose 𝑎=7.

Hence, the value of 𝑎 that makes the function 𝑓 continuous on its domain is 7.

Let us finish by recapping a few important concepts from this explainer.

Key Points

  • We say that a function is continuous on an interval (or a set) if it is continuous at each point in the interval (or the set).
  • Loosely speaking, a function is continuous on an interval if we can draw the graph of a continuous function over the interval without lifting our pencil from the paper.
  • The following types of functions are continuous everywhere:
    • a constant function,
    • a polynomial function,
    • an exponential function,
    • the absolute value function,
    • an odd root function (e.g., cube root),
    • the sine and cosine functions.
  • The following types of functions are continuous on their domains:
    • a rational function,
    • a logarithmic function,
    • an even root function (e.g., square root),
    • the tangent function.
  • Let 𝑓(𝑥) and 𝑔(𝑥) be continuous on sets 𝑆 and 𝑇 respectively. Then,
    • functions 𝑓(𝑥)+𝑔(𝑥), 𝑓(𝑥)𝑔(𝑥), and 𝑓(𝑥)𝑔(𝑥) are continuous on 𝑆𝑇,
    • 𝑓(𝑥)𝑔(𝑥) is continuous on the set 𝑆𝑇𝐴, where 𝐴 is the set 𝐴={𝑥𝑔(𝑥)=0},
    • 𝑓𝑔(𝑥) is continuous on the set 𝐵𝑇, where 𝐵 is the set 𝐵={𝑥𝑔(𝑥)𝑆}.
  • If the subfunctions of a piecewise-defined function are continuous, we only need to examine the continuity of the piecewise-defined function on the endpoints of the disjoint intervals.

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