Continuous Functions: Definition, Properties and Theorems

Continuity of a function at a point

Let \[ f:A\to\mathbb{R} \] be a function, and let \(x_0\in A\).

Definition. The function \(f\) is said to be continuous at \(x_0\) if, for every \(\varepsilon>0\), there exists \(\delta>0\) such that, for every \(x\in A\),

\[ |x-x_0|<\delta \quad\Longrightarrow\quad |f(x)-f(x_0)|<\varepsilon. \]

The meaning of the definition is precise. The number \(\varepsilon\) prescribes how close we wish \(f(x)\) to be to \(f(x_0)\); once any \(\varepsilon>0\) has been chosen, we must be able to determine a number \(\delta>0\) such that every \(x\) in the domain sufficiently close to \(x_0\) produces a value \(f(x)\) differing from \(f(x_0)\) by less than \(\varepsilon\).

In other words, the distance

\[ |f(x)-f(x_0)| \]

can be made as small as we please by choosing \(x\) sufficiently close to \(x_0\).

Continuity and limits

If \(x_0\) is a limit point of the domain \(A\), the continuity of \(f\) at \(x_0\) is equivalent to the condition

\[ \lim_{x\to x_0}f(x)=f(x_0). \]

This equality expresses the fundamental link between limit and continuity. The limit describes the behaviour of the values \(f(x)\) as \(x\) approaches \(x_0\); continuity requires that this behaviour agree with the value the function actually takes at the point.

Hence, when \(x_0\) is a limit point of the domain, three conditions must be satisfied in order to verify continuity by means of limits:

• \(f(x_0)\) must be defined;
• the limit \[ \lim_{x\to x_0}f(x) \] must exist;
• the limit must coincide with the value taken by the function: \[ \lim_{x\to x_0}f(x)=f(x_0). \]

If even one of these conditions fails to hold, \(f\) is not continuous at \(x_0\).

Example. Consider the function

\[ f(x)=x^2. \]

At the point \(x_0=2\) we have

\[ f(2)=4. \]

Moreover,

\[ \lim_{x\to2}x^2=4. \]

Hence

\[ \lim_{x\to2}f(x)=f(2), \]

and therefore \(f(x)=x^2\) is continuous at \(x_0=2\).

Geometric interpretation. Continuity at \(x_0\) expresses the fact that the point \((x_0,f(x_0))\) is consistent with the behaviour of the values \(f(x)\) as \(x\) approaches \(x_0\). In particular, the value taken by the function at the point coincides with the value towards which the values of the function tend in its vicinity.

The various ways in which this property can fail will be classified in the section devoted to points of discontinuity.

Continuity from the right and from the left

In some cases it is necessary to study separately the behaviour of the function as \(x\) approaches a point only from the right or only from the left. This leads to the notions of continuity from the right and continuity from the left.

Let

\[ f:A\to\mathbb{R} \]

and let \(x_0\in A\).

If \(x_0\) is a limit point of \(A\) from the right, \(f\) is said to be continuous from the right at \(x_0\) if

\[ \lim_{x\to x_0^+}f(x)=f(x_0). \]

Likewise, if \(x_0\) is a limit point of \(A\) from the left, \(f\) is said to be continuous from the left at \(x_0\) if

\[ \lim_{x\to x_0^-}f(x)=f(x_0). \]

When \(x_0\) is a limit point of the domain both from the left and from the right, \(f\) is continuous at \(x_0\) if and only if it is continuous both from the left and from the right at that same point.

Equivalently,

\[ \lim_{x\to x_0^-}f(x) = \lim_{x\to x_0^+}f(x) = f(x_0). \]

Example with a piecewise-defined function

Consider

\[ f(x)= \begin{cases} x+1, & x<1,\\ 2, & x=1,\\ 3-x, & x>1. \end{cases} \]

At the point \(x_0=1\),

\[ f(1)=2. \]

From the left we have

\[ \lim_{x\to1^-}f(x) = \lim_{x\to1^-}(x+1) = 2, \]

while from the right

\[ \lim_{x\to1^+}f(x) = \lim_{x\to1^+}(3-x) = 2. \]

Hence

\[ \lim_{x\to1^-}f(x) = \lim_{x\to1^+}f(x) = f(1) = 2, \]

and therefore \(f\) is continuous at \(x_0=1\).

Continuity at the endpoints of an interval

One-sided continuity is also essential at the endpoints of intervals.

If a function is defined on \([a,b]\), continuity from the right is considered at the point \(a\):

\[ \lim_{x\to a^+}f(x)=f(a), \]

whereas continuity from the left is considered at the point \(b\):

\[ \lim_{x\to b^-}f(x)=f(b). \]

Continuity on an interval

Having defined continuity at a single point, we may naturally extend the concept to an entire interval.

Open interval

A function \(f\) is said to be continuous on an open interval \(I\) if it is continuous at every point \(x_0\in I\).

For instance, \(f\) is continuous on \((a,b)\) if, for every \(x_0\in(a,b)\),

\[ \lim_{x\to x_0}f(x)=f(x_0). \]

Closed interval

In the case of a closed interval \([a,b]\), the approach at the endpoints must be considered while remaining within the interval. For this reason one-sided limits come into play.

A function

\[ f:[a,b]\to\mathbb{R} \]

is said to be continuous on \([a,b]\) if:

• it is continuous at every point \(x_0\in(a,b)\);
• it is continuous from the right at \(a\), that is, \[ \lim_{x\to a^+}f(x)=f(a); \]
• it is continuous from the left at \(b\), that is, \[ \lim_{x\to b^-}f(x)=f(b). \]

Half-open intervals

Similarly, if the domain is a half-open interval, the corresponding one-sided continuity is required at the endpoint belonging to the domain.

A function

\[ f:[a,b)\to\mathbb{R} \]

is continuous on \([a,b)\) if it is continuous at every point of \((a,b)\) and is continuous from the right at \(a\).

Likewise,

\[ f:(a,b]\to\mathbb{R} \]

is continuous on \((a,b]\) if it is continuous at every point of \((a,b)\) and is continuous from the left at \(b\).

Continuity on the domain

More generally, a function

\[ f:A\to\mathbb{R} \]

is said to be continuous on \(A\) if it is continuous at every point of its own domain, continuity being understood with respect to the points belonging to \(A\).

For example,

\[ f(x)=x^2 \]

is continuous on the whole of \(\mathbb{R}\), since for every \(x_0\in\mathbb{R}\)

\[ \lim_{x\to x_0}x^2=x_0^2=f(x_0). \]

The function

\[ f(x)=\frac{1}{x} \]

is, on the other hand, continuous on its own domain

\[ \mathbb{R}\setminus\{0\}. \]

The fact that it is not defined at \(x=0\) does not, therefore, prevent us from stating that it is continuous on its own domain. Continuity is a property that is verified at the points where the function is defined.

Continuity on an interval will play a fundamental role in the results that follow: the existence of zeros, the attainment of intermediate values, and the existence of absolute maxima and minima all depend on continuity over suitable intervals.

Operations with continuous functions

Continuity is preserved under the principal algebraic operations. This makes it possible to construct new continuous functions from functions whose continuity is already known.

Let \(f\) and \(g\) be continuous at a point \(x_0\). Then \(f+g\), \(f-g\), and \(f\cdot g\) are also continuous at \(x_0\).

For every constant \(c\in\mathbb{R}\), the function

\[ cf \]

is likewise continuous at \(x_0\).

If

\[ g(x_0)\neq0, \]

the quotient function

\[ \frac{f}{g} \]

is also continuous at \(x_0\).

Why these properties hold

Since \(f\) and \(g\) are continuous at \(x_0\),

\[ \lim_{x\to x_0}f(x)=f(x_0), \qquad \lim_{x\to x_0}g(x)=g(x_0). \]

For the sum, it follows from the properties of limits that

\[ \begin{aligned} \lim_{x\to x_0}[f(x)+g(x)] &= \lim_{x\to x_0}f(x) + \lim_{x\to x_0}g(x)\\ &= f(x_0)+g(x_0)\\ &= (f+g)(x_0). \end{aligned} \]

Hence \(f+g\) is continuous at \(x_0\). The results concerning the difference, the product, and the product by a constant are proved in an entirely analogous way.

For the quotient, if \(g(x_0)\neq0\),

\[ \lim_{x\to x_0}\frac{f(x)}{g(x)} = \frac{f(x_0)}{g(x_0)} = \left(\frac{f}{g}\right)(x_0), \]

and hence \(\displaystyle\frac{f}{g}\) is continuous at \(x_0\).

The same properties hold on an interval. If \(f\) and \(g\) are continuous on an interval \(I\), then \(f+g\), \(f-g\), \(f\cdot g\), and \(cf\) are continuous on \(I\). The quotient \(\displaystyle\frac{f}{g}\) is continuous at every point of \(I\) at which \(g\) does not vanish.

Composition of continuous functions

Another fundamental property concerns composition.

Let \(g\) be continuous at \(x_0\) and let \(f\) be continuous at \(g(x_0)\). Then the composite function

\[ (f\circ g)(x)=f(g(x)) \]

is continuous at \(x_0\).

Indeed,

\[ \lim_{x\to x_0}g(x)=g(x_0). \]

Since \(f\) is continuous at \(g(x_0)\),

\[ \lim_{x\to x_0}f(g(x)) = f\left(\lim_{x\to x_0}g(x)\right) = f(g(x_0)). \]

Hence

\[ \lim_{x\to x_0}(f\circ g)(x) = (f\circ g)(x_0), \]

and therefore \(f\circ g\) is continuous at \(x_0\).

Example

Consider

\[ g(x)=x^2+1 \]

and

\[ f(t)=\sqrt{t}. \]

Both functions are continuous on their respective domains. Moreover \(x^2+1>0\) for every \(x\in\mathbb{R}\), so the composition

\[ (f\circ g)(x)=\sqrt{x^2+1} \]

is continuous on the whole of \(\mathbb{R}\).

These results make it possible to recognise the continuity of a great many functions without having to apply the \(\varepsilon\)–\(\delta\) definition directly each time.

Continuity of elementary functions

The principal elementary functions are continuous at every point of their domain. Together with the properties of operations and composition, this result makes it possible to establish quickly the continuity of a wide class of functions.

Polynomial and rational functions

Every polynomial function

\[ P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 \]

is continuous on the whole of \(\mathbb{R}\).

A rational function

\[ f(x)=\frac{P(x)}{Q(x)} \]

is continuous at every point at which the denominator does not vanish, that is, on its own domain

\[ \{x\in\mathbb{R}:Q(x)\neq0\}. \]

Powers and roots

The usual power functions and root functions, considered on their respective real domains, are continuous.

For example,

\[ f(x)=\sqrt{x} \]

is continuous on

\[ [0,+\infty), \]

whereas

\[ f(x)=\sqrt[3]{x} \]

is continuous on the whole of \(\mathbb{R}\).

Exponential and logarithmic functions

The exponential function

\[ f(x)=a^x, \qquad a>0, \quad a\neq1, \]

is continuous for every \(x\in\mathbb{R}\).

The logarithmic function

\[ f(x)=\log_a(x), \qquad a>0, \quad a\neq1, \]

is continuous on its own domain

\[ (0,+\infty). \]

In particular, the natural logarithm

\[ f(x)=\ln(x) \]

is continuous for every \(x>0\).

Trigonometric functions

The functions

\[ \sin(x), \qquad \cos(x) \]

are continuous on the whole of \(\mathbb{R}\).

The tangent function

\[ \tan(x)=\frac{\sin(x)}{\cos(x)} \]

is continuous at the points where \(\cos(x)\neq0\), that is, for

\[ x\neq\frac{\pi}{2}+k\pi, \qquad k\in\mathbb{Z}. \]

Likewise, the other trigonometric functions defined by means of quotients are continuous at every point of their domain.

Absolute value

The function

\[ f(x)=|x| \]

is likewise continuous on the whole of \(\mathbb{R}\).

A more elaborate example

Consider

\[ f(x)=\frac{\sqrt{x^2+1}}{x^2+4}. \]

The polynomial \(x^2+1\) is continuous and satisfies

\[ x^2+1>0 \]

for every \(x\in\mathbb{R}\). Consequently,

\[ \sqrt{x^2+1} \]

is continuous on \(\mathbb{R}\).

Moreover,

\[ x^2+4>0 \]

for every \(x\in\mathbb{R}\), so the denominator never vanishes.

Therefore

\[ f(x)=\frac{\sqrt{x^2+1}}{x^2+4} \]

is continuous on the whole of \(\mathbb{R}\).

In general, when studying the continuity of an elementary function it is often enough to determine its domain correctly and to recognise the operations and compositions by which the function is built up.

Points of discontinuity

When a function is not continuous at a point of its domain, one speaks of a discontinuity. In the study of limits the same terminology is commonly used also for limit points of the domain at which the function is not defined, as occurs, for example, in removable discontinuities.

Suppose that \(x_0\) is a limit point of the domain both from the left and from the right. To describe the behaviour of the function at \(x_0\), the two one-sided limits are considered:

\[ \lim_{x\to x_0^-}f(x) \qquad\text{and}\qquad \lim_{x\to x_0^+}f(x). \]

Their behaviour makes it possible to distinguish several types of discontinuity.

Removable discontinuity

There is a removable discontinuity at \(x_0\) when the limit

\[ \lim_{x\to x_0}f(x)=L, \]

exists and is finite, but the function is either not defined at \(x_0\), or is defined there with

\[ f(x_0)\neq L. \]

The discontinuity can be removed by defining, or redefining,

\[ f(x_0)=L. \]

Example. Consider

\[ f(x)=\frac{x^2-1}{x-1}, \qquad x\neq1. \]

For \(x\neq1\),

\[ f(x) = \frac{(x-1)(x+1)}{x-1} = x+1. \]

Hence

\[ \lim_{x\to1}f(x)=2. \]

The function is not defined at \(x=1\), but by setting

\[ f(1)=2 \]

one obtains a function that is continuous at that point as well.

Discontinuity of the first kind, or jump discontinuity

There is a discontinuity of the first kind, also called a jump discontinuity, when both one-sided limits exist and are finite but take different values:

\[ \lim_{x\to x_0^-}f(x)=L_1, \qquad \lim_{x\to x_0^+}f(x)=L_2, \qquad L_1\neq L_2. \]

In this case the limit as \(x\to x_0\) does not exist. The quantity

\[ |L_2-L_1| \]

is called the size of the jump.

Example. Consider

\[ f(x)= \begin{cases} 0, & x<0,\\ 1, & x\geq0. \end{cases} \]

We have

\[ \lim_{x\to0^-}f(x)=0, \qquad \lim_{x\to0^+}f(x)=1. \]

The two one-sided limits are finite but different. Hence \(x=0\) is a point of discontinuity of the first kind.

Discontinuity of the second kind

There is a discontinuity of the second kind when at least one of the two one-sided limits fails to exist finitely.

This may occur either because at least one of the limits is infinite, or because at least one of them fails to exist even as an infinite limit.

First example. Consider

\[ f(x)=\frac{1}{x}. \]

As \(x\to0\),

\[ \lim_{x\to0^-}\frac{1}{x}=-\infty, \qquad \lim_{x\to0^+}\frac{1}{x}=+\infty. \]

Since the two one-sided limits are infinite, \(x=0\) is a point of discontinuity of the second kind. Moreover,

\[ x=0 \]

is a vertical asymptote of the graph.

Second example. Consider

\[ f(x)=\sin\left(\frac{1}{x}\right), \qquad x\neq0. \]

As \(x\to0\), the argument \(\displaystyle\frac{1}{x}\) grows without bound in absolute value, and the function continues to oscillate between \(-1\) and \(1\).

Consequently, neither the left-hand limit nor the right-hand limit exists, not even as an infinite limit. Thus \(x=0\) is likewise a point of discontinuity of the second kind.

The various types of discontinuity show that continuity can fail in profoundly different ways. The study of one-sided limits makes it possible to describe precisely the local behaviour of the function at the problematic points.

The Sign Preservation Theorem for continuous functions

A fundamental consequence of continuity is the Sign Preservation Theorem: if a continuous function takes a strictly positive or strictly negative value at a point, then it retains the same sign throughout a suitable neighbourhood of that point.

Let

\[ f:A\to\mathbb{R} \]

be continuous at \(x_0\in A\).

If

\[ f(x_0)>0, \]

then there exists \(\delta>0\) such that, for every \(x\in A\),

\[ |x-x_0|<\delta \quad\Longrightarrow\quad f(x)>0. \]

Likewise, if

\[ f(x_0)<0, \]

there exists \(\delta>0\) such that

\[ |x-x_0|<\delta \quad\Longrightarrow\quad f(x)<0. \]

Proof. Suppose

\[ f(x_0)>0. \]

Choose

\[ \varepsilon=\frac{f(x_0)}{2}>0. \]

Since \(f\) is continuous at \(x_0\), there exists \(\delta>0\) such that

\[ |x-x_0|<\delta \quad\Longrightarrow\quad |f(x)-f(x_0)| < \frac{f(x_0)}{2}. \]

From the properties of the absolute value it follows that

\[ -\frac{f(x_0)}{2} < f(x)-f(x_0) < \frac{f(x_0)}{2}. \]

Considering the left-hand inequality,

\[ f(x) > f(x_0)-\frac{f(x_0)}{2} = \frac{f(x_0)}{2} > 0. \]

Hence \(f\) retains a positive sign throughout a suitable neighbourhood of \(x_0\). The case \(f(x_0)<0\) is proved analogously.

Why the hypothesis \(f(x_0)\neq0\) is essential

If

\[ f(x_0)=0, \]

continuity alone does not make it possible to determine the sign of the function in a neighbourhood of \(x_0\).

For example,

\[ f(x)=x \]

is continuous at \(x_0=0\), yet it takes negative values to the left of \(0\) and positive values to the right.

The Sign Preservation Theorem will play a fundamental role in Bolzano's Theorem, in which continuity will make it possible to deduce the existence of at least one zero when a function takes values of opposite sign at the endpoints of an interval.

Bolzano's Theorem and the Intermediate Value Theorem

Continuity on an interval produces a fundamental global consequence: a continuous function cannot pass from one value to another without also taking every intermediate value.

Bolzano's Theorem

Let

\[ f:[a,b]\to\mathbb{R} \]

be continuous on \([a,b]\).

If

\[ f(a)f(b)<0, \]

that is, if \(f(a)\) and \(f(b)\) are of opposite sign, then there exists at least one point

\[ c\in(a,b) \]

such that

\[ f(c)=0. \]

The theorem thus guarantees the existence of at least one zero of the function within the interval, but it does not assert that such a zero is unique.

Geometrically, if the graph of a continuous function lies on one side of the \(x\)-axis at one endpoint of the interval and on the other side at the other endpoint, it must necessarily cross the \(x\)-axis at least once.

Example. Consider

\[ f(x)=x^3+x-1 \]

on the interval \([0,1]\).

Being a polynomial, \(f\) is continuous on the whole of \(\mathbb{R}\), and hence also on \([0,1]\). Moreover,

\[ f(0)=-1, \qquad f(1)=1. \]

Therefore

\[ f(0)f(1)=-1<0. \]

By Bolzano's Theorem there exists at least one point

\[ c\in(0,1) \]

such that

\[ f(c)=0. \]

The theorem thus makes it possible to prove the existence of a solution without necessarily determining its exact value.

The Intermediate Value Theorem

Bolzano's Theorem is a particular case of a more general result: the Intermediate Value Theorem.

Let

\[ f:[a,b]\to\mathbb{R} \]

be continuous on \([a,b]\).

For every real number \(k\) lying between \(f(a)\) and \(f(b)\), there exists at least one point \(c\in[a,b]\) such that

\[ f(c)=k. \]

In particular, if

\[ f(a)<k<f(b) \]

or

\[ f(b)<k<f(a), \]

then there exists at least one point

\[ c\in(a,b) \]

such that

\[ f(c)=k. \]

This means that the image of an interval under a continuous function has no gaps: between any two values taken by the function, every intermediate value is also attained.

For example, if

\[ f(a)=2, \qquad f(b)=7, \]

then, for every \(k\) such that

\[ 2<k<7, \]

there exists at least one \(c\in(a,b)\) with

\[ f(c)=k. \]

The theorem does not guarantee uniqueness: the same intermediate value may be attained at more than one point of the interval.

Bolzano's Theorem as a particular case

If \(f(a)\) and \(f(b)\) are of opposite sign, then \(0\) lies between \(f(a)\) and \(f(b)\).

Applying the Intermediate Value Theorem with

\[ k=0, \]

immediately yields a point \(c\in(a,b)\) such that

\[ f(c)=0. \]

Bolzano's Theorem may therefore be regarded as the application of the Intermediate Value Theorem to the particular value \(k=0\).

The hypothesis of continuity is essential. A discontinuous function may take values of opposite sign at the endpoints of an interval without taking the value \(0\) anywhere inside it. It is precisely continuity that prevents the image of the function from exhibiting a gap between the values considered.

Weierstrass's Theorem

Another fundamental consequence of continuity concerns the existence of absolute maxima and minima.

Let

\[ f:[a,b]\to\mathbb{R} \]

be a function continuous on the closed and bounded interval \([a,b]\).

Weierstrass's Theorem states that \(f\) attains both a minimum value and a maximum value on \([a,b]\).

In other words, there exist points

\[ x_m,x_M\in[a,b] \]

such that

\[ f(x_m)\leq f(x)\leq f(x_M) \]

for every \(x\in[a,b]\).

The value

\[ f(x_m) \]

is the absolute minimum of \(f\) on \([a,b]\), while

\[ f(x_M) \]

is the absolute maximum.

It is important to distinguish the points at which the extreme values are attained from the extreme values themselves: \(x_m\) and \(x_M\) belong to the domain, while \(f(x_m)\) and \(f(x_M)\) are, respectively, the minimum and the maximum of the function.

Boundedness and attainment of the extreme values

The theorem guarantees, in particular, that a function continuous on a closed and bounded interval is bounded.

Setting

\[ m=\inf f([a,b]), \qquad M=\sup f([a,b]), \]

we have

\[ m\leq f(x)\leq M \]

for every \(x\in[a,b]\).

Weierstrass's Theorem asserts, moreover, something stronger: these two extreme values are actually attained. Indeed, there exist \(x_m,x_M\in[a,b]\) such that

\[ m=f(x_m), \qquad M=f(x_M). \]

Example. Consider

\[ f(x)=x^2-2x+3 \]

on the interval \([0,3]\).

Being a polynomial, \(f\) is continuous on the whole of \(\mathbb{R}\), and hence also on \([0,3]\). Weierstrass's Theorem therefore guarantees the existence of both an absolute minimum and an absolute maximum.

We write

\[ f(x)=(x-1)^2+2. \]

Since

\[ (x-1)^2\geq0, \]

the minimum value is attained at \(x=1\):

\[ f(1)=2. \]

Moreover, for \(x\in[0,3]\),

\[ |x-1|\leq2, \]

so

\[ (x-1)^2\leq4. \]

Hence

\[ f(x)=(x-1)^2+2\leq6, \]

with equality at \(x=3\). The absolute maximum is therefore

\[ f(3)=6. \]

Why the hypotheses matter

The hypotheses of the theorem are essential. Continuity alone does not suffice to guarantee the existence of an absolute maximum and minimum if the interval is not closed and bounded.

For example,

\[ f(x)=x \]

is continuous on the open interval \((0,1)\), but attains neither its infimum \(0\) nor its supremum \(1\). Consequently, it possesses on \((0,1)\) neither an absolute minimum nor an absolute maximum.

Likewise,

\[ f(x)=x \]

is continuous on the whole of \(\mathbb{R}\), but is unbounded and therefore possesses neither an absolute maximum nor an absolute minimum.

Interpretation by means of compactness

In \(\mathbb{R}\), every closed and bounded set is compact. The interval \([a,b]\) is therefore compact.

Since the continuous image of a compact set is again compact,

\[ f([a,b]) \]

is a compact subset of \(\mathbb{R}\).

Consequently, \(f([a,b])\) is closed and bounded. Being bounded, it possesses an infimum and a supremum; being closed, it contains both of these values. They are therefore actually attained by the function and constitute, respectively, the absolute minimum and the absolute maximum.

Weierstrass's Theorem is one of the central results of mathematical analysis and will play a fundamental role in the study of maximum and minimum problems.

Continuity of the inverse function

Continuity is preserved, under suitable hypotheses, in passing to the inverse function as well.

Let \(I\subseteq\mathbb{R}\) be an interval, and let

\[ f:I\to\mathbb{R} \]

be a function that is continuous and strictly monotonic on \(I\).

Since \(f\) is strictly monotonic, it is injective, and hence admits an inverse function defined on the image

\[ J=f(I). \]

The theorem on the continuity of the inverse function states that

\[ f^{-1}:J\to I \]

is continuous on \(J\).

Moreover, \(J=f(I)\) is itself an interval. This follows from the Intermediate Value Theorem: since \(f\) is continuous on an interval, every value lying between any two values taken by \(f\) is likewise attained.

The inverse function also preserves the type of monotonicity:

• if \(f\) is strictly increasing, then \(f^{-1}\) is likewise strictly increasing;
• if \(f\) is strictly decreasing, then \(f^{-1}\) is likewise strictly decreasing.

Geometrically, the graph of the inverse function is obtained by reflecting the graph of \(f\) in the line

\[ y=x. \]

Example: the cubic function

Consider

\[ f(x)=x^3, \qquad x\in\mathbb{R}. \]

The function is continuous and strictly increasing on \(\mathbb{R}\). Moreover,

\[ f(\mathbb{R})=\mathbb{R}. \]

Its inverse function is

\[ f^{-1}(x)=\sqrt[3]{x}. \]

By the preceding theorem, \(f^{-1}\) is continuous on the whole of \(\mathbb{R}\).

Example: the exponential function

Consider

\[ f(x)=a^x, \qquad a>0, \quad a\neq1. \]

The exponential function is continuous and strictly monotonic on \(\mathbb{R}\), with image

\[ (0,+\infty). \]

Its inverse function is the logarithm

\[ f^{-1}(x)=\log_a(x), \]

which is therefore continuous on

\[ (0,+\infty). \]

The continuity of the inverse function completes the picture of the principal properties of continuous functions: local continuity and continuity on intervals, preservation under operations and composition, the classification of discontinuities, and the main global consequences.