Ever heard of a function being described as continuous in the past? These are the functions with graphs that do not contain holes, asymptotes, and gaps between curves. These “nice” graphs we’ve encountered in the past are called continuous functions.

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*Continuous functions are functions that **look smooth throughout, and we can graph them without lifting our own pens. *

We can also assess a function’s continuity through limits and higher maths – and that’s our focus in this article.

We’ll learn about the conditions of continuous functions.We’ll apply our techniques in evaluating limits to confirm if a function is continuous.Apply graphical techniques as well to identify whether a graph is continuous or not.In Calculus, we’ll also encounter continuous functions again, so learning about them now can be helpful, especially for those about to progress to differential calculus soon. Why don’t we go ahead and understand what these functions represent?

## What is a continuous function?

Continuous functions are functions that have no restrictions throughout their domain or a given interval. Their graphs won’t contain any asymptotes or signs of discontinuities as well.

The graph of $f(x) = x^3 – 4x^2 – x + 10$ as shown below is a great example of a continuous function’s graph. As can be seen, the graph extends throughout both the $x$-axis’ positive and negative sides.

The graph above is an example of a function that is not continuous because of a discontinuity at $x = 3$. Observe how as $x$ approaches from the left of $3$, the function approaches $-\infty$? Similarly, when $x$ approaches from the right of $3$, the function approaches $\infty$.

This means that the function’s limit does not exist and consequently, making it not continuous. This is also why the **other name for this discontinuity is infinite discontinuity**.

Keep in mind that a function may contain more than one discontinuity, so better double-check your function’s graph or limits well.

Now that we’ve dealt with the possible conditions where a function may not be continuous, why don’t we go ahead and learn more of other important properties of continuous functions?

## What are other important properties of continuous functions?

We’ve now learned about identifying continuous functions and being able to assess functions for discontinuities. These properties below will help us confirm and prove if a function is continuous by using simpler functions.

**Property 1: **$\boldsymbol{k \cdot f(x)}$

When $k$ is a constant and $f(x)$ is a continuous function when $x = a$, then $k\cdot f(x)$ is also continuous at $ x= a$.

**Property 2: **$\boldsymbol{f(x) + g(x)}$

When $f(x)$ and $g(x)$ are both continuous functions when $x = a$, then the resulting function when we add $f(x)$ and $g(x)$ will also be continuous at $x =a$.

**Property 3: **$\boldsymbol{f(x) – g(x)}$

When $f(x)$ and $g(x)$ are both continuous functions when $x = a$, then the difference between the functions will also be a function that is continuous when $x = a$.

**Property 4: **$\boldsymbol{f(x) \cdot g(x)}$

If the functions $f(x)$ and $g(x)$ are continuous functions at $x = a$, the two functions’ product is also continuous at $x = a$.

**Property 5: **$\boldsymbol{\dfrac{f(x)}{g(x)}}$

If the functions $f(x)$ and $g(x)$ are continuous when $x = a$ and $g(a) \neq 0$, the ratio of $f(x)$ and $g(x)$, is also continuous at $x = a$.

**Property 6: **$\boldsymbol{f(g(x))}$

When $g(x)$ is continuous at $x = a$ and $f(x)$ is continuous at $x = g(a)$, then $f(g(x))$ will also be continuous at $x = a$.

### Common functions that are continuous

Here are some of the functions that you may have encountered in the past that are known to be continuous within their domain.

Sine function: $y = \sin x$Cosine function: $y = \cos x$Tangent function: $y = \tan x$Radical function: $y = \sqrt{x}$Exponential function: $y = a^x$, where $a > 0$ and $y = e^x$Natural logarithmic function: $y = \ln x$You might wonder, tangent and radical functions have restrictions, so how are they continuous? This is when it’s important to highlight that these functions are continuous only **within their domain**.

For example, $\sqrt{x – 1}$ has a domain of $<0, \infty)$, so it’s expected to be continuous within the interval, $<0, \infty)$. Meaning, if asked whether $\sqrt{x – 1}$ is continuous at $x = -2$, of course, it won’t be since the function is not defined at $x = -2$.

This can help you identify continuous functions (along with the properties discussed above), especially when it has complex expressions.

Let’s go ahead and try out more examples to understand continuous and discontinuous functions better.

*Example 1*

Fill in the blanks to make the following statements true.

a. If $f(2) = -24$ and $f(x)$ is a continuous function, $\lim_{x \rightarrow 2} f(x)$ is equal to ____________.**b. If $g(x)$ has a hole at $(5, 4)$, the function is not continuous at ____________.c. If h(x) contains a vertical asymptote at $x = -1$, the function is not continuous when _________.**

**Solution**

**Recall that when the function and its limit are defined at $x = a$, the third condition will require the two’s values to be equal.**

**a. This means that for $f(x)$ to be continuous, $\lim_{x \rightarrow 2} f(x)$ must also be equal to $\boldsymbol{-24}$.When a function has a hole at $(a, b)$, there is removable discontinuity at $x = a$.b. Since $g(x)$ has a hole at $(5, 4)$, it has a removable discontinuity at the $x$-coordinate of the hole. Meaning, $g(x)$ is not continuous at $\boldsymbol{x = 5}$.When a function contains a vertical asymptote, its value and limit will be undefined at the vertical asymptote’s value. c. Using this information, $h(x)$ is continuous throughout its domain except it’s equal to $-1$. Hence, the function is continuous when $\boldsymbol{x = -1}$.**

*Example 2*

**Discuss the continuity of the following function at the given corresponding points.**

**a. $f(x) = -4x^2 + 8$, when $x = 4$b. $g(x) = \dfrac{5x + 1}{2x – 3}$, when $x = 3$c. $h(x) = \sqrt{x^2 + 2}$, when $x = -2$**

**Solution**

**When confirming whether a function is continuous, make sure to check three things:**

**The function is defined at $ x= a$.The limit exists when the function approaches $a$.Last, if we have $f(x)$, $\lim_{x \rightarrow a} f(x) = f(a)$.**

Let’s start with $f(x) = -4x^2 + 8$ and see if this function satisfies all three conditions.

Since $f(x)$ is a polynomial, all values of $x$, including $4$ is defined at $f(x)$. In fact, $f(4)$ is equal to $-4(4)^2 + 8 = -56$.For polynomial functions, $\lim_{x \rightarrow a} f(x) = f(a)$, so $\lim_{x \rightarrow 4} f(4) = -56$.This also means that $\lim_{x \rightarrow 4} f(x) = f(4)$.As we have discussed in the previous sections, all polynomials are continuous.

a. All these confirm that $\boldsymbol{f(x)}$ **is a continuous function**.

We can inspect the second function, $g(x) = \dfrac{5x + 1}{2x – 3}$, using the same process. Let’s begin by evaluating $g(3)$ as shown below.

$\begin{aligned}g(3)&=\dfrac{5(3) + 1}{2(3) – 3}\\&=\dfrac{16}{0}\end{aligned}$

From this, we can see that $g(3)$ is not defined. We don’t need to check the remaining conditions when this happens.

b. Since $g(3)$ is not defined, $\boldsymbol{g(x)}$ **is not a continuous function**.

Let’s move on to the third function – $ h(x) = \sqrt{x^2 + 2}$.

Substitute $x = -2$ into the expression for $h(x)$. Hence, we have $ h(-2) = \sqrt{((-2)^2 + 2} = \sqrt{6}$. This means that $h(x)$ is defined at $x = -2$.Next, let’s evaluate the limit of $h(x)$ as it approaches $x = -2$: $\lim_{x \rightarrow -2} \sqrt{x^2 + 2} = \sqrt{(-2)^2 + 2} = \sqrt{6}$.Comparing the limit and value of the $h(x)$ when $x = -2$, we have $h(-2) = \lim_{x \rightarrow -2} h(x) = \sqrt{6}$.c. Seeing that $h(x)$ satisfies all three conditions when $x = -2$, $\boldsymbol{h(x)}$ **is a continuous function**.

*Example 3*

Is the function, $ f(x)=\left\{\begin{matrix}-3x + 1,&xSolution

As we have done in the previous example, we can check for continuity by reviewing the three conditions.

Starting by checking if $f(a)$ is defined. When $x \geq 4$, we have $f(x)$ = 2x – 5. This means that $f(4) = 2(4) – 5 = 3$.

Let’s go ahead and observe the limit of $f(x)$ as it approaches $4$. We’re working with a piecewise function, so it’s best to check the one-sided limits of $f(x)$.

$\lim_{x \rightarrow 4^{-}} f(x)$ | $\lim_{x \rightarrow 4^{+}} f(x)$ |

$\begin{aligned}\lim_{x \rightarrow 4^{-}} f(x) &= \lim_{x \rightarrow 4} -3x + 1\\&=-3(4) + 1\\&=-11\end{aligned}$ | $\begin{aligned}\lim_{x \rightarrow 4^{+}} f(x) &= \lim_{x \rightarrow 4} 2x – 5\\&=2(4) – 5\\&=3\end{aligned}$ |

From this, we can see that $\lim_{x \rightarrow 4^{-}} f(x) \neq \lim_{x \rightarrow 4^{+}} f(x)$, so the limit for $f(x)$ is not defined.

At this point, since $f(x)$ does not meet the second condition, $f(x)$ **is not continuous**.

*Example 4*

Identify which of the following functions are discontinuous. If working with a discontinuous function, identify the type of discontinuity it has.

a. $f(x) = -2x^3 + 5x – 9$b. $f(x) = \dfrac{1}{4x^2 + 4}$c. $f(x) = \dfrac{2x^2 – 2x}{4x}$d. $f(x) = \dfrac{x – 4}{x^2 – 6x + 8}$

Solution

a. Since the function, $f(x) = -2x^3 + 5x – 9$, is a polynomial function, it is **continuous** throughout its domain, $(-\infty, \infty)$.

Although the function is rational and may contain asymptotes, the denominator of $f(x)$ is $4x^2 + 4$, which can never be negative.

b. This means that $f(x) = \dfrac{1}{4x^2 + 4}$ has no restrictions and when this happens, this rational function is **continuous.**

For $f(x) = \dfrac{2x^2 – 2x}{4x}$, let’s factor the numerator first and see if the numerator and denominator share a common factor.

$\begin{aligned}\dfrac{2x^2 – 2x}{4x}&= \dfrac{2x(x – 2)}{2(2x)}\\&= \dfrac{\cancel{2x}(x – 2)}{2\cancel{(2x)}}\\&=\dfrac{x-2}{2}\end{aligned}$

Since $f(x)$’s numerator and denominator share a common factor of $2x$, so it has a hole at $x = 0$.

To find the hole’s $y$-coordinate, we can substitute $x = 0$ into the simplified form of $f(x)$.

$\begin{aligned}f(x) &= \dfrac{x-2}{2}\\ f(0)&= \dfrac{0-2}{2}\\&=-1\end{aligned}$

c. Since $f(x)$ has a hole and consequently, a discontinuity at $x = 0$. Since we have a discontinuity at a hole, $f(x)$ **is not continuous **and we can consider it a **removable discontinuity**.

Let’s go ahead and express the denominator of $f(x)$ in factored form: $\dfrac{x – 4}{x^2 – 6x + 8} = \dfrac{x – 4}{(x -2)(x – 4)}$.

Since $x – 4$ is a common factor shared by $f(x)$’s numerator and denominator, there is a hole at $x = 4$. Find the $y$-coordinate by substituting $x=4$ into the simplified form of $f(x)$.

$\begin{aligned}f(x)&= \dfrac{\cancel{x – 4}}{(x -2)\cancel{(x – 4)}}\\&= \dfrac{1}{x -2}\\\\f(4)&= \dfrac{1}{4 -2}\\&= \dfrac{1}{2}\end{aligned}$

From the simplified form of $f(x)$, $\dfrac{1}{x – 2}$, we can see that $f(x)$ will also have a vertical asymptote at $x = 2$.

d. This means that $f(x)$ **is not continuous **and $x = 4$ is a **removable discontinuity** while $x =2$ is an **infinite discontinuity**.

*Example 5*

Given that the function, $ f(x)=\left\{\begin{matrix}Mx + N,&x\leq -1\\ 3x^2 – 5Mx -N,&-1 1\end{matrix}\right.$, is continuous for all values of $x$, find the values of $M$ and $N$.

Solution

Let’s inspect the continuity of $f(x)$ in each of the piecewise function’s components.

When $x \leq -1$, $f(x) = Mx +N$, and since this is a polynomial function, we can say that $Mx + N$ will always be continuous regardless of $M$ and $N$’s values.The same reasoning applies when $f(x) = 3x^2 – 5x – N$ and $f(x) = -6$ for the intervals, $-1 -1$, respectively.To make sure that $f(x)$ is continuous throughout, we’ll need to inspect how it behaves at $x = -1$ and $x = 1$.

Starting with the first condition of continuous functions, $f(-1)$ and $f(1)$ must be defined.

$f(-1)$ | $\begin{aligned}f(x) &= Mx + N, \text{when } x \leq -1\\f(-1)&=M(-1) + N\\&=N -M \end{aligned}$ |

$f(1)$ | $\begin{aligned}f(x) &= 3x^2 – 5Mx -N, \text{when } -1 |

This means that $f(-1) = N -M $ and $f(1) = 3 – 5M – N $ must be defined for $f(x)$ to be continuous.

Let’s evaluate the one-sided limits of $f(x)$ as $x$ approaches both $1$ and $-1$. Starting with $\lim_{x \rightarrow -1^{-}} f(x)$ and $\lim_{x \rightarrow -1^{+}} f(x)$:

$\lim_{x \rightarrow -1^{-}} f(x)$ | $\begin{aligned}\lim_{x \rightarrow -1^{-}} f(x) &= \lim_{x \rightarrow -1 } Mx + N \\&=M(-1) + N\\&= N – M\end{aligned}$ |

$\lim_{x \rightarrow -1^{+}} f(x)$ | $\begin{aligned}\lim_{x \rightarrow -1^{+}} f(x) &= \lim_{x \rightarrow -1 } 3x^2 – 5Mx -N\\&= 3(-1)^2 -5M(-1)- N\\&=3 + 5M -N\end{aligned}$ |

For the limit of $f(x)$ to exist and be defined, both one-sided limits must be equal to each other. Equate the two one-sided limits’ expressions to each other.

$\begin{aligned}\lim_{x \rightarrow -1^{-}} f(x) &= \lim_{x \rightarrow -1^{+}} f(x) \\N -M &=3 + 5M -N\\2N – 6M &= 3\end{aligned}$

We apply the same process to observe the one-sided limits of $f(x)$ as it approaches $x = 1$.

$\lim_{x \rightarrow 1^{-}} f(x)$ | $\begin{aligned}\lim_{x \rightarrow 1^{-}} f(x) &= \lim_{x \rightarrow 1 } 3x^2 – 5Mx -N \\&= 3(1)^2 – 5M(1) – N\\&= 3 – 5M – N\end{aligned}$ |

$\lim_{x \rightarrow 1^{+}} f(x)$ | $\begin{aligned}\lim_{x \rightarrow 1^{-}} f(x) &= \lim_{x \rightarrow 1 } -6 \\&=-6\end{aligned}$ |

Equating the two limits, we have the equation shown below.

$\begin{aligned}\lim_{x \rightarrow 1^{-}} f(x) &= \lim_{x \rightarrow 1^{+}} f(x)\\3 – 5M – N &= -6\\-5M – N &= -9\\ 5M + N&=9\end{aligned}$

This means that $M$ and $N$ must satisfy the simultaneous equations, $2N – 6M = 3$ and $5M + N = 9$. Let’s apply what we’ve in learned in solving systems of linear equations to find $M$ and $N$.

Isolate $N$ from the second equation.Substitute this expression into the first equation to solve for $M$.Use the value of $M$ to find $N$.$\begin{aligned}5M + N &= 9\\N&= 9- 5M\\\\2N – 6M &= 3\\2(9 – 5M) – 6M &= 3\\18 – 10M- 6M&=3\\18 – 16M &= 3\\-16M &= -15\\ M&= \dfrac{15}{16}\end{aligned}$

Using $M = \dfrac{15}{16}$, we can now find $N$ using $N = 9-5M$.

$\begin{aligned}N&= 9- 5\left(\dfrac{15}{16}\right)\\&=9 – \dfrac{75}{16}\\&= \dfrac{69}{16}\end{aligned}$

This means that for $f(x)$ to be continuous, we need $M$ and $N$, to be equal to $\dfrac{15}{16}$ and $\dfrac{69}{16}$, respectively.

**Practice Questions **

**Practice Questions**

1. Fill in the blanks to make the following statements true.

a. If $f(4) = -\dfrac{1}{2}$ and $f(x)$ is a continuous function, $\lim_{x \rightarrow 4} f(x)$ is equal to ____________.b. If $g(x)$ has a hole at $(-2, -1)$, the function is not continuous at ____________.c. If $h(x)$ contains a vertical asymptote at $x = \sqrt{3}$, the function is not continuous when _________.

2. Discuss the continuity of the following function at the given corresponding points. a. $f(x) = 2x^2 – 3x + 14$, when $x = -1$b. $g(x) = \dfrac{-2x + 3}{4x – 1}$, when $x = \dfrac{1}{4}$c. $h(x) = \sqrt{4x^2 + 1}$, when $x = -\dfrac{1}{2}$

3. Is the function, $ f(x)=\left\{\begin{matrix}-5x + 3,&x4. Identify which of the following functions are discontinuous. If working with a discontinuous function, identify the type of discontinuity it has.a. $f(x) = 4x^3 – 12x^2 + 6x – 20$b. $f(x) = \dfrac{1}{2x^2 + 1}$c. $f(x) = \dfrac{x}{3x^2 – 6x}$d. $f(x) = \dfrac{x + 5}{x^2 +10x + 25}$5. Given that the function, $ f(x)=\left\{\begin{matrix}Mx + N,&x\leq -1\\ -2x^2 +6 Mx -N,&-1 1\end{matrix}\right.$, is continuous for all values of $x$, find the values of $M$ and $N$.

**Answer Key**

1.

a. $-\dfrac{1}{2}$

b. $x=-2$

c. $x=\sqrt{3}$

2.

a. Continuous

b. Not continuous

c. Continuous

3. Not continuous

4.

a. $f(x)$ is continuous.

b. $f(x)$ is continuous.

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c. $f(x)$ is not continuous ; removable discontinuity at $x=0$ and an infinite discontinuity at $x=2$.