Ever heard of a function being defined as constant in the past? These space the attributes with graphs that perform not save on computer holes, asymptotes, and also gaps in between curves. This “nice” graphs we’ve encountered in the previous are called continuous functions.

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

We can likewise assess a function’s continually through limits and higher maths – and that’s our emphasis in this article.

We’ll learn about the conditions of constant functions.We’ll apply our methods in examining limits to check if a function is continuous.Apply graphical techniques also to determine whether a graph is consistent or not.

In Calculus, we’ll also encounter consistent functions again, so learning about them now deserve to be helpful, particularly for those about to progress to differential calculus soon. Why don’t us go ahead and understand what these attributes represent?

What is a constant function?

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

The graph that $f(x) = x^3 – 4x^2 – x + 10$ together shown listed below is a great example of a consistent function’s graph. As have the right to be seen, the graph extends throughout both the $x$-axis’ optimistic and negative sides.


The graph over is an example of a function that is not continuous because of a discontinuity at $x = 3$. Observe how as $x$ philosophies from the left that $3$, the role approaches $-\\infty$? Similarly, as soon as $x$ ideologies from the appropriate of $3$, the role approaches $\\infty$.

This means that the function’s limit does no exist and consequently, do it no continuous. This is additionally why the other surname for this discontinuity is boundless discontinuity.

Keep in mind the a function may contain an ext than one discontinuity, so much better double-check her function’s graph or limits well.

Now the we’ve dealt with the possible conditions where a duty may no be continuous, why don’t us go ahead and learn much more of other important properties of continuous functions?

What are other important properties of consistent functions?

We’ve currently learned around identifying continuous functions and being able come assess functions for discontinuities. This properties listed below will assist us confirm and also prove if a duty is continuous by using less complicated functions.

Property 1: $\\boldsymbolk \\cdot f(x)$

When $k$ is a consistent and $f(x)$ is a continuous function as soon as $x = a$, climate $k\\cdot f(x)$ is also constant at $ x= a$.

Property 2: $\\boldsymbolf(x) + g(x)$

When $f(x)$ and also $g(x)$ room both continuous functions as soon as $x = a$, climate the resulting function when we include $f(x)$ and $g(x)$ will also be consistent at $x =a$.

Property 3: $\\boldsymbolf(x) – g(x)$

When $f(x)$ and $g(x)$ space both consistent functions once $x = a$, then the difference between the attributes will likewise be a role that is consistent when $x = a$.

Property 4: $\\boldsymbolf(x) \\cdot g(x)$

If the features $f(x)$ and also $g(x)$ are continuous functions in ~ $x = a$, the 2 functions’ product is also constant at $x = a$.

Property 5: $\\boldsymbol\\dfracf(x)g(x)$

If the features $f(x)$ and $g(x)$ are constant when $x = a$ and $g(a) \\neq 0$, the proportion of $f(x)$ and also $g(x)$, is also constant at $x = a$.

Property 6: $\\boldsymbolf(g(x))$

When $g(x)$ is constant at $x = a$ and also $f(x)$ is consistent at $x = g(a)$, climate $f(g(x))$ will also be consistent at $x = a$.

Common features that space continuous

Here are several of the features that girlfriend may have encountered in the previous that are well-known to be continuous within their domain.

Sine function: $y = \\sin x$Cosine function: $y = \\cos x$Tangent function: $y = \\tan x$Radical function: $y = \\sqrtx$Exponential function: $y = a^x$, whereby $a > 0$ and $y = e^x$Natural logarithmic function: $y = \\ln x$

You can wonder, tangent and also radical functions have restrictions, so how are they continuous? This is once it’s vital to highlight that these features are consistent only within their domain.

For example, $\\sqrtx – 1$ has a domain that $<0, \\infty)$, for this reason it’s meant to be continuous within the interval, $<0, \\infty)$. Meaning, if asked even if it is $\\sqrtx – 1$ is continuous at $x = -2$, the course, it won’t be due to the fact that the function is not characterized at $x = -2$.

This can assist you identify continuous functions (along through the properties debated above), especially when that has complicated expressions.

Let’s go ahead and shot out much more examples to understand constant and discontinuous features better.

Example 1

Fill in the blanks to make the complying with statements true.

a. If $f(2) = -24$ and also $f(x)$ is a continuous function, $\\lim_x \\rightarrow 2 f(x)$ is equal to ____________.b. If $g(x)$ has a hole at $(5, 4)$, the duty is not continuous at ____________.c. If h(x) consists of a upright asymptote in ~ $x = -1$, the role is not continuous when _________.


Recall that as soon as the function and its border are characterized at $x = a$, the 3rd condition will require the two’s values to it is in equal.

a. This method that because that $f(x)$ to it is in continuous, $\\lim_x \\rightarrow 2 f(x)$ must additionally be same to $\\boldsymbol-24$.When a role has a feet at $(a, b)$, there is removable discontinuity in ~ $x = a$.b. Due to the fact that $g(x)$ has a feet at $(5, 4)$, it has a removable discontinuity at the $x$-coordinate that the hole. Meaning, $g(x)$ is not consistent at $\\boldsymbolx = 5$.When a function contains a upright asymptote, the value and also limit will certainly be undefined at the vertical asymptote’s value. C. Making use of this information, $h(x)$ is consistent throughout that is domain other than it’s same to $-1$. Hence, the function is continuous when $\\boldsymbolx = -1$.

Example 2

Discuss the continuity of the following role at the given equivalent points.

a. $f(x) = -4x^2 + 8$, when $x = 4$b. $g(x) = \\dfrac5x + 12x – 3$, when $x = 3$c. $h(x) = \\sqrtx^2 + 2$, as soon as $x = -2$


When confirming even if it is a function is continuous, make certain to check three things:

The duty is identified at $ x= a$.The limit exists as soon as the role approaches $a$.Last, if we have actually $f(x)$, $\\lim_x \\rightarrow a f(x) = f(a)$.

Let’s begin with $f(x) = -4x^2 + 8$ and also see if this role satisfies all 3 conditions.

Since $f(x)$ is a polynomial, all worths of $x$, including $4$ is identified at $f(x)$. In fact, $f(4)$ is same 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 disputed in the vault sections, every polynomials space continuous.

a. Every these confirm that $\\boldsymbolf(x)$ is a constant function.

 We can inspect the second function, $g(x) = \\dfrac5x + 12x – 3$, utilizing the very same process. Let’s begin by assessing $g(3)$ as displayed below.

$\\beginalignedg(3)&=\\dfrac5(3) + 12(3) – 3\\\\&=\\dfrac160\\endaligned$

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

b. Due to the fact that $g(3)$ is not defined, $\\boldsymbolg(x)$ is no a continuous function.

Let’s relocate on to the third duty – $ h(x) = \\sqrtx^2 + 2$.

Substitute $x = -2$ into the expression because that $h(x)$. Hence, we have actually $ h(-2) = \\sqrt((-2)^2 + 2 = \\sqrt6$. This means that $h(x)$ is defined at $x = -2$.Next, let’s evaluate the border of $h(x)$ as it approaches $x = -2$: $\\lim_x \\rightarrow -2 \\sqrtx^2 + 2 = \\sqrt(-2)^2 + 2 = \\sqrt6$.Comparing the limit and value of the $h(x)$ once $x = -2$, we have actually $h(-2) = \\lim_x \\rightarrow -2 h(x) = \\sqrt6$.

c. Seeing that $h(x)$ satisfies every three conditions when $x = -2$, $\\boldsymbolh(x)$ is a consistent function.

Example 3

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

As we have done in the previous example, we can examine for continuity by reviewing the 3 conditions.

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

Let’s go ahead and also observe the border of $f(x)$ as it approaches $4$. We’re working through a piecewise function, for this reason it’s best to check the one-sided borders of $f(x)$.

$\\lim_x \\rightarrow 4^- f(x)$

$\\lim_x \\rightarrow 4^+ f(x)$

$\\beginaligned\\lim_x \\rightarrow 4^- f(x) &= \\lim_x \\rightarrow 4 -3x + 1\\\\&=-3(4) + 1\\\\&=-11\\endaligned$

$\\beginaligned\\lim_x \\rightarrow 4^+ f(x) &= \\lim_x \\rightarrow 4 2x – 5\\\\&=2(4) – 5\\\\&=3\\endaligned$

From this, we have the right to see that $\\lim_x \\rightarrow 4^- f(x) \\neq \\lim_x \\rightarrow 4^+ f(x)$, so the border for $f(x)$ is not defined.

At this point, because $f(x)$ does not satisfy the second condition, $f(x)$ is no continuous.

Example 4

Identify which of the following attributes are discontinuous. If working through a discontinuous function, identify the form of discontinuity the has.

a. $f(x) = -2x^3 + 5x – 9$b. $f(x) = \\dfrac14x^2 + 4$c. $f(x) = \\dfrac2x^2 – 2x4x$d. $f(x) = \\dfracx – 4x^2 – 6x + 8$


a. Due to the fact that the function, $f(x) = -2x^3 + 5x – 9$, is a polynomial function, that is continuous transparent its domain, $(-\\infty, \\infty)$.

Although the role is rational and may save on computer asymptotes, the denominator that $f(x)$ is $4x^2 + 4$, which can never it is in negative.

b. This means that $f(x) = \\dfrac14x^2 + 4$ has actually no restrictions and when this happens, this rational role is continuous.

For $f(x) = \\dfrac2x^2 – 2x4x$, let’s factor the numerator first and view if the numerator and also denominator re-superstructure a typical factor.

$\\beginaligned\\dfrac2x^2 – 2x4x&= \\dfrac2x(x – 2)2(2x)\\\\&= \\dfrac\\cancel2x(x – 2)2\\cancel(2x)\\\\&=\\dfracx-22\\endaligned$

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

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

$\\beginalignedf(x) &= \\dfracx-22\\\\ f(0)&= \\dfrac0-22\\\\&=-1\\endaligned$

c. Since $f(x)$ has actually a hole and also consequently, a discontinuity at $x = 0$. Due to the fact that we have actually a discontinuity in ~ a hole, $f(x)$ is not consistent and we can think about it a removable discontinuity.

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

Since $x – 4$ is a usual factor shared by $f(x)$’s numerator and denominator, there is a feet at $x = 4$. Uncover the $y$-coordinate by substituting $x=4$ into the simplified type of $f(x)$.

$\\beginalignedf(x)&= \\dfrac\\cancelx – 4(x -2)\\cancel(x – 4)\\\\&= \\dfrac1x -2\\\\\\\\f(4)&= \\dfrac14 -2\\\\&= \\dfrac12\\endaligned$

From the simplified form of $f(x)$, $\\dfrac1x – 2$, we can see that $f(x)$ will also have a vertical asymptote in ~ $x = 2$.

d. This way that $f(x)$ is not consistent and $x = 4$ is a removable discontinuity if $x =2$ is one infinite discontinuity.

Example 5

Given that the function, $ f(x)=\\left\\{\\beginmatrixMx + N,&x\\leq -1\\\\ 3x^2 – 5Mx -N,&-1 1\\endmatrix\\right.$, is constant for all values of $x$, uncover the worths of $M$ and also $N$.


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

When $x \\leq -1$, $f(x) = Mx +N$, and also since this is a polynomial function, we deserve to say that $Mx + N$ will always be constant regardless that $M$ and also $N$’s values.The exact same reasoning uses when $f(x) = 3x^2 – 5x – N$ and $f(x) = -6$ because that the intervals, $-1 -1$, respectively.

To make certain that $f(x)$ is continuous throughout, we’ll need to check how the behaves at $x = -1$ and also $x = 1$.

Starting through the an initial condition of consistent functions, $f(-1)$ and also $f(1)$ should be defined.


$\\beginalignedf(x) &= Mx + N, \\textwhen x \\leq -1\\\\f(-1)&=M(-1) + N\\\\&=N -M \\endaligned$


$\\beginalignedf(x) &= 3x^2 – 5Mx -N, \\textwhen -1

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

Let’s evaluate the one-sided boundaries of $f(x)$ as $x$ philosophies both $1$ and $-1$. Starting with $\\lim_x \\rightarrow -1^- f(x)$ and also $\\lim_x \\rightarrow -1^+ f(x)$:

$\\lim_x \\rightarrow -1^- f(x)$

$\\beginaligned\\lim_x \\rightarrow -1^- f(x) &= \\lim_x \\rightarrow -1 Mx + N \\\\&=M(-1) + N\\\\&= N – M\\endaligned$

$\\lim_x \\rightarrow -1^+ f(x)$

$\\beginaligned\\lim_x \\rightarrow -1^+ f(x) &= \\lim_x \\rightarrow -1 3x^2 – 5Mx -N\\\\&= 3(-1)^2 -5M(-1)- N\\\\&=3 + 5M -N\\endaligned$

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

$\\beginaligned\\lim_x \\rightarrow -1^- f(x) &= \\lim_x \\rightarrow -1^+ f(x) \\\\N -M &=3 + 5M -N\\\\2N – 6M &= 3\\endaligned$

We apply the same procedure to observe the one-sided borders of $f(x)$ as it philosophies $x = 1$.

$\\lim_x \\rightarrow 1^- f(x)$

$\\beginaligned\\lim_x \\rightarrow 1^- f(x) &= \\lim_x \\rightarrow 1 3x^2 – 5Mx -N \\\\&= 3(1)^2 – 5M(1) – N\\\\&= 3 – 5M – N\\endaligned$

$\\lim_x \\rightarrow 1^+ f(x)$

$\\beginaligned\\lim_x \\rightarrow 1^- f(x) &= \\lim_x \\rightarrow 1 -6 \\\\&=-6\\endaligned$

Equating the two limits, we have actually the equation presented below.

$\\beginaligned\\lim_x \\rightarrow 1^- f(x) &= \\lim_x \\rightarrow 1^+ f(x)\\\\3 – 5M – N &= -6\\\\-5M – N &= -9\\\\ 5M + N&=9\\endaligned$

This way that $M$ and also $N$ must fulfill the simultaneously equations, $2N – 6M = 3$ and $5M + N = 9$. Let’s apply what we’ve in learned in solving systems of direct equations to find $M$ and also $N$.

Isolate $N$ indigenous the 2nd equation.Substitute this expression right into the very first equation to fix for $M$.Use the value of $M$ to find $N$.

$\\beginaligned5M + N &= 9\\\\N&= 9- 5M\\\\\\\\2N – 6M &= 3\\\\2(9 – 5M) – 6M &= 3\\\\18 – 10M- 6M&=3\\\\18 – 16M &= 3\\\\-16M &= -15\\\\ M&= \\dfrac1516\\endaligned$

Using $M = \\dfrac1516$, we have the right to now discover $N$ making use of $N = 9-5M$.

$\\beginalignedN&= 9- 5\\left(\\dfrac1516\\right)\\\\&=9 – \\dfrac7516\\\\&= \\dfrac6916\\endaligned$

This way that for $f(x)$ to it is in continuous, we require $M$ and also $N$, come be same to $\\dfrac1516$ and $\\dfrac6916$, respectively.

Practice inquiries

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

a. If $f(4) = -\\dfrac12$ and $f(x)$ is a constant function, $\\lim_x \\rightarrow 4 f(x)$ is same to ____________.b. If $g(x)$ has actually a feet at $(-2, -1)$, the duty is not continuous at ____________.c. If $h(x)$ consists of a vertical asymptote in ~ $x = \\sqrt3$, the role is not continuous when _________.

2. Comment on the continuous of the following role at the given matching points. A. $f(x) = 2x^2 – 3x + 14$, when $x = -1$b. $g(x) = \\dfrac-2x + 34x – 1$, as soon as $x = \\dfrac14$c. $h(x) = \\sqrt4x^2 + 1$, when $x = -\\dfrac12$

3. Is the function, $ f(x)=\\left\\{\\beginmatrix-5x + 3,&x4. Identify which of the following features are discontinuous. If working v a discontinuous function, determine the type of discontinuity that has.a. $f(x) = 4x^3 – 12x^2 + 6x – 20$b. $f(x) = \\dfrac12x^2 + 1$c. $f(x) = \\dfracx3x^2 – 6x$d. $f(x) = \\dfracx + 5x^2 +10x + 25$5. Given that the function, $ f(x)=\\left\\{\\beginmatrixMx + N,&x\\leq -1\\\\ -2x^2 +6 Mx -N,&-1 1\\endmatrix\\right.$, is continuous for all worths of $x$, uncover the values of $M$ and also $N$.

Answer Key


a. $-\\dfrac12$

b. $x=-2$

c. $x=\\sqrt3$


a. Continuous

b. No continuous

c. Continuous

3. No continuous


a. $f(x)$ is continuous.

b. $f(x)$ is continuous.

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c. $f(x)$ is not constant ; removable discontinuity in ~ $x=0$ and an boundless discontinuity in ~ $x=2$.