In this collection we present 20 practice problems on continuous functions, arranged in order of increasing difficulty and solved step by step, with a detailed explanation of every stage of the argument.
The problems provide practice in applying the principal ideas of the theory: verifying continuity at a point and on an interval, continuity from the right and from the left, the study of piecewise-defined functions, the determination of parameters ensuring continuity, operations on and composition of continuous functions, and the classification of discontinuities.
The final section deals with applications of the principal theorems concerning continuous functions, among them the permanence-of-sign theorem, Bolzano's theorem, the intermediate value theorem, Weierstrass's theorem, and the theorem on the continuity of the inverse function.
The aim is to learn not merely to determine whether a function is continuous, but above all to recognise which properties and which theorems are to be invoked in each type of problem, justifying every conclusion rigorously.
Problem 1 — level ★☆☆☆☆
Verify that the function
\[ f(x)=x^2-3x+4 \]
is continuous at the point \(x_0=2\).
Answer
The function is continuous at \(x_0=2\).
Solution
The function
\[ f(x)=x^2-3x+4 \]
is a polynomial function. Every polynomial function is continuous on the whole of \(\mathbb{R}\), and hence, in particular, at the point \(x_0=2\).
We may nonetheless verify explicitly the condition linking continuity to the limit. We first compute the value taken by the function at the point:
\[ f(2)=2^2-3\cdot2+4=4-6+4=2. \]
Since polynomials are continuous functions, the limit as \(x\to2\) coincides with the value taken by the function:
\[ \lim_{x\to2}(x^2-3x+4)=2. \]
We therefore have
\[ \lim_{x\to2}f(x)=2=f(2). \]
The condition
\[ \lim_{x\to2}f(x)=f(2) \]
is thus satisfied, and \(f\) is continuous at \(x_0=2\).
Problem 2 — level ★☆☆☆☆
Determine at which points the function
\[ f(x)=\frac{x+1}{x-3} \]
is continuous.
Answer
The function is continuous on its domain
\[ \mathbb{R}\setminus\{3\}. \]
Solution
The function is the quotient of the two polynomials
\[ P(x)=x+1 \]
and
\[ Q(x)=x-3. \]
Polynomials are continuous functions on the whole of \(\mathbb{R}\). The quotient of two continuous functions is itself a continuous function at every point at which the denominator does not vanish.
Let us determine, then, where the denominator equals zero:
\[ x-3=0, \]
so that
\[ x=3. \]
The function is not defined at \(x=3\), and its domain is accordingly
\[ \mathbb{R}\setminus\{3\}. \]
At every point \(x_0\neq3\) the denominator takes a non-zero value, so that the quotient is continuous there.
We conclude that
\[ f(x)=\frac{x+1}{x-3} \]
is continuous at every point of its domain, that is, on
\[ (-\infty,3)\cup(3,+\infty). \]
Problem 3 — level ★☆☆☆☆
Verify that the function
\[ f(x)=\sqrt{x} \]
is continuous at the endpoint \(x_0=0\) of its domain.
Answer
The function is continuous from the right at \(x_0=0\), and is therefore continuous at \(0\) relative to its domain.
Solution
The function
\[ f(x)=\sqrt{x} \]
has domain
\[ [0,+\infty). \]
The point \(0\) is the left-hand endpoint of the domain, so we cannot approach it through points of the domain lying to its left.
To verify continuity at this endpoint we must accordingly consider the limit from the right.
The value of the function at the point is
\[ f(0)=\sqrt{0}=0. \]
Moreover,
\[ \lim_{x\to0^+}\sqrt{x}=0, \]
so that
\[ \lim_{x\to0^+}f(x)=f(0). \]
The function is thus continuous from the right at \(0\). Since continuity is to be understood relative to the points belonging to the domain, we may assert that \(f(x)=\sqrt{x}\) is continuous also at the endpoint \(x=0\) of its domain.
Problem 4 — level ★★☆☆☆
Determine the value of the parameter \(k\in\mathbb{R}\) for which the function
\[ f(x)= \begin{cases} x^2+1, & x<1,\\ k, & x=1,\\ 2x, & x>1 \end{cases} \]
is continuous at \(x_0=1\).
Answer
The function is continuous at \(x_0=1\) if and only if
\[ k=2. \]
Solution
For the function to be continuous at \(x_0=1\), the left-hand limit, the right-hand limit, and the value \(f(1)\) must all coincide.
Let us first examine the limit from the left. For \(x<1\), the function is given by
\[ f(x)=x^2+1. \]
Being a polynomial, this expression is continuous, so that
\[ \lim_{x\to1^-}f(x) = \lim_{x\to1^-}(x^2+1) = 1^2+1 = 2. \]
For \(x>1\), on the other hand,
\[ f(x)=2x. \]
This too is a polynomial function, whence
\[ \lim_{x\to1^+}f(x) = \lim_{x\to1^+}2x = 2. \]
The two limits coincide:
\[ \lim_{x\to1^-}f(x) = \lim_{x\to1^+}f(x) = 2. \]
The limit
\[ \lim_{x\to1}f(x)=2 \]
therefore exists. At the point \(x=1\), however, the function takes the value
\[ f(1)=k. \]
Continuity requires that
\[ \lim_{x\to1}f(x)=f(1), \]
and hence that
\[ 2=k. \]
We conclude that the only value rendering the function continuous at \(x=1\) is
\[ k=2. \]
Problem 5 — level ★★☆☆☆
Determine the value of the parameter \(a\in\mathbb{R}\) for which the function
\[ f(x)= \begin{cases} ax+1, & x\leq2,\\ x^2-a, & x>2 \end{cases} \]
is continuous at \(x_0=2\).
Answer
The function is continuous at \(x_0=2\) for
\[ a=1. \]
Solution
The function changes its defining expression at the point \(x=2\), so we must compare its behaviour from the left and from the right.
For \(x\leq2\),
\[ f(x)=ax+1. \]
The value of the function at the point \(2\) is thus
\[ f(2)=2a+1, \]
and from the same expression we obtain the left-hand limit:
\[ \lim_{x\to2^-}f(x)=2a+1. \]
For \(x>2\), on the other hand,
\[ f(x)=x^2-a, \]
and hence
\[ \lim_{x\to2^+}f(x)=4-a. \]
For the limit as \(x\to2\) to exist, the left-hand and right-hand limits must coincide:
\[ 2a+1=4-a. \]
Solving this equation,
\[ 2a+a=4-1, \]
that is,
\[ 3a=3, \]
whence
\[ a=1. \]
Let us verify this. For \(a=1\),
\[ f(2)=2\cdot1+1=3, \]
while
\[ \lim_{x\to2^+}f(x)=4-1=3. \]
Hence
\[ \lim_{x\to2^-}f(x) = \lim_{x\to2^+}f(x) = f(2) = 3, \]
and the function is accordingly continuous at \(x=2\) for \(a=1\).
Problem 6 — level ★★☆☆☆
Study the continuity at \(x_0=2\) of the function
\[ f(x)= \begin{cases} \displaystyle\frac{x^2-4}{x-2}, & x\neq2,\\ 5, & x=2. \end{cases} \]
Classify the resulting discontinuity, if any, and state how it may be removed.
Answer
At \(x=2\) the function has a removable discontinuity. It is removed by redefining
\[ f(2)=4. \]
Solution
To study the continuity at \(x=2\) we must compare the limit of the function with the value it takes at the point.
For \(x\neq2\) we may factorise the numerator:
\[ x^2-4=(x-2)(x+2). \]
Hence, for \(x\neq2\),
\[ f(x) = \frac{(x-2)(x+2)}{x-2} = x+2. \]
This simplification is valid in computing the limit, since we are there considering values of \(x\) close to \(2\) but distinct from it.
Consequently,
\[ \lim_{x\to2}f(x) = \lim_{x\to2}(x+2) = 4. \]
The limit exists and is finite. The function, however, has been defined at the point \(x=2\) by
\[ f(2)=5. \]
We therefore have
\[ \lim_{x\to2}f(x)=4\neq5=f(2), \]
so that the function fails to be continuous at \(x=2\).
Since the limit exists and is finite yet does not coincide with the value taken by the function at the point, this is a removable discontinuity.
To remove it, it suffices to redefine the value of the function by setting
\[ f(2)=4. \]
With this new definition we have
\[ \lim_{x\to2}f(x)=f(2)=4, \]
and the redefined function is therefore continuous at \(x=2\).
Problem 7 — level ★★☆☆☆
Study and classify the discontinuity of the function
\[ f(x)= \begin{cases} x+1, & x<0,\\ x^2+2, & x\geq0 \end{cases} \]
at the point \(x_0=0\).
Answer
At \(x=0\) the function has a discontinuity of the first kind, or jump discontinuity, of magnitude \(1\).
Solution
Since the function changes its defining expression at \(x=0\), we compute the left-hand and right-hand limits separately.
For \(x<0\),
\[ f(x)=x+1, \]
so that
\[ \lim_{x\to0^-}f(x) = \lim_{x\to0^-}(x+1) = 1. \]
For \(x\geq0\),
\[ f(x)=x^2+2, \]
and hence
\[ \lim_{x\to0^+}f(x) = \lim_{x\to0^+}(x^2+2) = 2. \]
Both the left-hand and right-hand limits exist and are finite, but they differ:
\[ 1\neq2. \]
Consequently the limit
\[ \lim_{x\to0}f(x) \]
does not exist.
When the left-hand and right-hand limits are both finite but unequal, one speaks of a discontinuity of the first kind, also called a jump discontinuity.
The magnitude of the jump is the absolute value of the difference between the two limits:
\[ |2-1|=1. \]
Hence \(x=0\) is a point of discontinuity of the first kind, with a jump of magnitude \(1\).
Problem 8 — level ★★☆☆☆
Study and classify the discontinuity of the function
\[ f(x)=\frac{x+1}{x-2} \]
at the point \(x_0=2\).
Answer
At \(x=2\) the function has a discontinuity of the second kind. Moreover, \(x=2\) is a vertical asymptote.
Solution
The function is not defined at \(x=2\), since the denominator vanishes there:
\[ 2-2=0. \]
Let us study the limits from the left and from the right.
As \(x\to2\), the numerator
\[ x+1 \]
tends to the positive value \(3\).
Consider now the denominator. If \(x\to2^-\), then
\[ x-2<0, \]
and its value approaches \(0\) while remaining negative. The quotient of a positive quantity close to \(3\) by a negative quantity tending to \(0\) decreases without bound, so that
\[ \lim_{x\to2^-}\frac{x+1}{x-2}=-\infty. \]
If instead \(x\to2^+\), then
\[ x-2>0, \]
and the denominator tends to \(0\) through positive values. It follows that
\[ \lim_{x\to2^+}\frac{x+1}{x-2}=+\infty. \]
At least one of the two one-sided limits fails to exist finitely; here, in fact, both are infinite.
By definition, \(x=2\) is therefore a point of discontinuity of the second kind.
Furthermore, since the function diverges in absolute value as \(x\) approaches \(2\), the line
\[ x=2 \]
is a vertical asymptote of the graph.
Problem 9 — level ★★★☆☆
Determine where the function
\[ f(x)=\sqrt{x^2+3} \]
is continuous, justifying the answer by means of the composition of continuous functions.
Answer
The function is continuous on the whole of \(\mathbb{R}\).
Solution
Consider the inner function
\[ g(x)=x^2+3 \]
and the outer function
\[ h(t)=\sqrt{t}. \]
The function \(g\) is a polynomial, and is therefore continuous on the whole of \(\mathbb{R}\).
Moreover, for every \(x\in\mathbb{R}\),
\[ x^2\geq0, \]
and consequently
\[ x^2+3\geq3>0. \]
The values taken by \(g\) therefore always lie in the domain of the square-root function.
The function
\[ h(t)=\sqrt{t} \]
is continuous on its domain \([0,+\infty)\), and in particular at every value \(g(x)=x^2+3\).
By the theorem on the composition of continuous functions, the composite function
\[ (h\circ g)(x)=h(g(x))=\sqrt{x^2+3} \]
is continuous for every \(x\in\mathbb{R}\).
We conclude that
\[ f(x)=\sqrt{x^2+3} \]
is continuous on the whole of \(\mathbb{R}\).
Problem 10 — level ★★★☆☆
Determine the domain and the intervals of continuity of the function
\[ f(x)=\ln(x^2-1). \]
Answer
The domain is
\[ (-\infty,-1)\cup(1,+\infty), \]
and the function is continuous throughout its domain.
Solution
The natural logarithm
\[ \ln(t) \]
is defined only when its argument is strictly positive.
We must therefore impose
\[ x^2-1>0. \]
Factorising,
\[ x^2-1=(x-1)(x+1). \]
The product is positive when the two factors have the same sign, which gives
\[ x<-1 \]
or
\[ x>1. \]
The domain is thus
\[ D=(-\infty,-1)\cup(1,+\infty). \]
Let us now consider continuity. The inner function
\[ g(x)=x^2-1 \]
is a polynomial, and is therefore continuous on the whole of \(\mathbb{R}\).
The outer function
\[ h(t)=\ln(t) \]
is continuous for \(t>0\).
At every point of the domain of \(f\), the argument \(x^2-1\) is positive, so that we may apply the theorem on the composition of continuous functions.
It follows that
\[ f(x)=\ln(x^2-1) \]
is continuous on each of the two intervals
\[ (-\infty,-1) \]
and
\[ (1,+\infty). \]
Problem 11 — level ★★★☆☆
Verify that the function
\[ f(x)=\sqrt{4-x^2} \]
is continuous on the closed interval \([-2,2]\), paying particular attention to the endpoints.
Answer
The function is continuous throughout the closed interval \([-2,2]\).
Solution
Let us first determine the domain. For the square root to be defined we require
\[ 4-x^2\geq0, \]
a condition equivalent to
\[ x^2\leq4, \]
that is,
\[ -2\leq x\leq2. \]
The domain is thus precisely
\[ [-2,2]. \]
The function
\[ g(x)=4-x^2 \]
is polynomial, and hence continuous. On the interval under consideration it satisfies
\[ g(x)\geq0. \]
Since the square-root function is continuous on its domain, the composite function
\[ f(x)=\sqrt{4-x^2} \]
is continuous at every interior point
\[ x_0\in(-2,2). \]
It remains to examine the endpoints of the interval.
At the left-hand endpoint \(x=-2\) we must consider the limit from the right. We have
\[ f(-2)=\sqrt{4-(-2)^2}=0 \]
and
\[ \lim_{x\to-2^+}\sqrt{4-x^2}=0=f(-2), \]
so that \(f\) is continuous from the right at \(-2\).
At the right-hand endpoint \(x=2\), on the other hand, we consider the limit from the left:
\[ f(2)=\sqrt{4-2^2}=0 \]
and
\[ \lim_{x\to2^-}\sqrt{4-x^2}=0=f(2), \]
so that the function is continuous from the left at \(2\).
Being continuous at every interior point and possessing the appropriate one-sided continuity at the two endpoints, \(f\) is continuous on the entire closed interval \([-2,2]\).
Problem 12 — level ★★★☆☆
Using the permanence-of-sign theorem, show that the function
\[ f(x)=x^2-3x+3 \]
is positive in a suitable neighbourhood of \(x_0=1\).
Answer
There exists \(\delta>0\) such that
\[ |x-1|<\delta \quad\Longrightarrow\quad f(x)>0. \]
Solution
The permanence-of-sign theorem states that if a function is continuous at a point and takes a strictly positive value there, then it retains a positive sign in a suitable neighbourhood of that point.
Let us verify the two required hypotheses.
The function
\[ f(x)=x^2-3x+3 \]
is a polynomial, and is consequently continuous on the whole of \(\mathbb{R}\), and in particular at \(x_0=1\).
Let us now compute the value taken at the point:
\[ f(1)=1^2-3\cdot1+3=1, \]
so that
\[ f(1)=1>0. \]
Both hypotheses of the permanence-of-sign theorem are thus satisfied: \(f\) is continuous at \(1\), and \(f(1)\) is strictly positive.
We may therefore conclude that there exists at least one number \(\delta>0\) such that, for every \(x\in\mathbb{R}\),
\[ |x-1|<\delta \quad\Longrightarrow\quad f(x)>0. \]
It is not necessary to determine the value of \(\delta\) explicitly: the theorem guarantees the existence of a neighbourhood of \(1\) throughout which the function retains the positive sign taken at the point.
Problem 13 — level ★★★☆☆
Prove, by means of Bolzano's theorem, that the equation
\[ x^3-2x-2=0 \]
has at least one solution in the interval \((1,2)\).
Answer
There exists at least one point
\[ c\in(1,2) \]
such that
\[ c^3-2c-2=0. \]
Solution
Consider the function
\[ f(x)=x^3-2x-2. \]
To apply Bolzano's theorem we must verify two conditions: that the function is continuous on the closed interval \([1,2]\), and that the values it takes at the endpoints are of opposite sign.
The function \(f\) is a polynomial, and is therefore continuous on the whole of \(\mathbb{R}\); in particular, it is continuous on \([1,2]\).
Let us compute the value at the left-hand endpoint:
\[ f(1)=1^3-2\cdot1-2=1-2-2=-3, \]
so that
\[ f(1)<0. \]
At the right-hand endpoint we obtain
\[ f(2)=2^3-2\cdot2-2=8-4-2=2, \]
whence
\[ f(2)>0. \]
The values \(f(1)\) and \(f(2)\) are of opposite sign; equivalently,
\[ f(1)f(2)=(-3)\cdot2=-6<0. \]
All the hypotheses of Bolzano's theorem are thus satisfied.
There exists, then, at least one point
\[ c\in(1,2) \]
such that
\[ f(c)=0. \]
This means precisely that the equation
\[ x^3-2x-2=0 \]
possesses at least one solution lying between \(1\) and \(2\).
Bolzano's theorem guarantees the existence of at least one solution, but does not by itself allow us to conclude that it is unique.
Problem 14 — level ★★★☆☆
Use Bolzano's theorem to prove that the equation
\[ 2^x=3 \]
has at least one solution in the interval \((1,2)\).
Answer
There exists at least one point
\[ c\in(1,2) \]
such that
\[ 2^c=3. \]
Solution
To bring the problem within the scope of Bolzano's theorem, we move every term to one side and consider the function
\[ f(x)=2^x-3. \]
The equation
\[ 2^x=3 \]
is equivalent to
\[ f(x)=0. \]
The exponential function \(2^x\) is continuous on the whole of \(\mathbb{R}\); the constant function \(3\) is likewise continuous, and the difference of continuous functions is continuous.
Hence
\[ f(x)=2^x-3 \]
is continuous on the whole of \(\mathbb{R}\), and therefore also on the closed interval \([1,2]\).
Let us compute the values at the endpoints:
\[ f(1)=2^1-3=-1, \]
while
\[ f(2)=2^2-3=1. \]
We thus have
\[ f(1)<0 \qquad\text{and}\qquad f(2)>0, \]
or, equivalently,
\[ f(1)f(2)=-1<0. \]
By Bolzano's theorem there exists at least one point
\[ c\in(1,2) \]
such that
\[ f(c)=0. \]
Since \(f(c)=2^c-3\), we obtain
\[ 2^c-3=0, \]
that is,
\[ 2^c=3. \]
The equation therefore possesses at least one solution in the interval \((1,2)\).
Problem 15 — level ★★★☆☆
Using the intermediate value theorem, prove that there exists at least one point
\[ c\in(1,2) \]
such that
\[ c^3=5. \]
Answer
There exists at least one point \(c\in(1,2)\) such that \(c^3=5\).
Solution
Consider the function
\[ f(x)=x^3 \]
on the closed interval \([1,2]\).
The function is polynomial, and is therefore continuous on the whole of \(\mathbb{R}\); in particular, it is continuous on \([1,2]\).
Let us compute the values taken at the endpoints:
\[ f(1)=1^3=1 \]
and
\[ f(2)=2^3=8. \]
The value we wish to attain is
\[ k=5. \]
We observe that
\[ 1<5<8, \]
that is,
\[ f(1)<5<f(2). \]
The number \(5\) thus lies between the values taken by the function at the endpoints of the interval.
By the intermediate value theorem, there exists at least one point
\[ c\in(1,2) \]
such that
\[ f(c)=5. \]
Since \(f(c)=c^3\), we obtain
\[ c^3=5. \]
The theorem thus guarantees the existence of the required point without our having to determine its value explicitly.
Problem 16 — level ★★★★☆
Consider the function
\[ f(x)=(x-2)^2+1 \]
on the interval \([0,5]\). Verify the hypotheses of Weierstrass's theorem and determine the absolute minimum and maximum of the function on the interval.
Answer
The absolute minimum is
\[ 1, \]
attained at \(x=2\), while the absolute maximum is
\[ 10, \]
attained at \(x=5\).
Solution
The function
\[ f(x)=(x-2)^2+1 \]
is a polynomial, and is therefore continuous on the whole of \(\mathbb{R}\); in particular, it is continuous on the closed and bounded interval
\[ [0,5]. \]
The hypotheses of Weierstrass's theorem are thus satisfied, and we may already conclude that \(f\) attains both an absolute minimum and an absolute maximum on the interval.
Let us now determine these values.
For every \(x\in[0,5]\),
\[ (x-2)^2\geq0, \]
and consequently
\[ f(x)=(x-2)^2+1\geq1. \]
Equality holds when
\[ (x-2)^2=0, \]
that is, when
\[ x=2. \]
Since \(2\in[0,5]\), the absolute minimum is
\[ f(2)=1. \]
To determine the maximum, observe that, for \(x\in[0,5]\), the distance of \(x\) from the point \(2\) cannot exceed \(3\); indeed,
\[ |x-2|\leq3, \]
with the value \(3\) attained at the endpoint \(x=5\).
Squaring,
\[ (x-2)^2\leq9, \]
so that
\[ f(x)=(x-2)^2+1\leq10. \]
For \(x=5\),
\[ f(5)=(5-2)^2+1=9+1=10. \]
The absolute maximum is thus \(10\), attained at \(x=5\).
In conclusion,
\[ \min_{x\in[0,5]}f(x)=1 \]
and
\[ \max_{x\in[0,5]}f(x)=10. \]
Problem 17 — level ★★★★☆
Consider the function
\[ f(x)=|x-1| \]
on the interval \([-2,3]\). Use Weierstrass's theorem to establish the existence of an absolute minimum and maximum, and then determine them.
Answer
The absolute minimum is \(0\), attained at \(x=1\). The absolute maximum is \(3\), attained at \(x=-2\).
Solution
The absolute-value function is continuous on the whole of \(\mathbb{R}\), and so too is the function
\[ x-1. \]
Hence
\[ f(x)=|x-1| \]
is continuous on the closed and bounded interval
\[ [-2,3]. \]
By Weierstrass's theorem, \(f\) certainly attains an absolute minimum and an absolute maximum on the interval.
The absolute value is always non-negative:
\[ |x-1|\geq0, \]
and it equals zero precisely when
\[ x-1=0, \]
that is, when
\[ x=1. \]
Since \(1\in[-2,3]\), the absolute minimum is
\[ f(1)=|1-1|=0. \]
To find the maximum we must instead identify, within the interval, the point furthest from \(1\), since \(f(x)=|x-1|\) represents precisely the distance of \(x\) from \(1\).
At the left-hand endpoint,
\[ f(-2)=|-2-1|=|-3|=3. \]
At the right-hand endpoint,
\[ f(3)=|3-1|=2. \]
On the interval \([-2,3]\), no point is further from \(1\) than the point \(-2\) is. Hence
\[ |x-1|\leq3 \]
for every \(x\in[-2,3]\).
The absolute maximum is therefore
\[ f(-2)=3. \]
Problem 18 — level ★★★★☆
Consider the function
\[ f(x)=x^3+1. \]
Determine its inverse function and establish whether the inverse is continuous.
Answer
The inverse function is
\[ f^{-1}(x)=\sqrt[3]{x-1}, \]
and is continuous on the whole of \(\mathbb{R}\).
Solution
The function
\[ f(x)=x^3+1 \]
is a polynomial, and is therefore continuous on the whole of \(\mathbb{R}\).
Furthermore, the cubic function \(x^3\) is strictly increasing on \(\mathbb{R}\); adding the constant \(1\) does not alter this monotonicity, so \(f\) too is strictly increasing.
Being strictly increasing, \(f\) is injective, and admits an inverse function on its image.
The function \(x^3\) takes every real value, and the addition of \(1\) does not affect the fact that the image is the whole of the real numbers. Hence
\[ f(\mathbb{R})=\mathbb{R}. \]
To find the inverse, set
\[ y=x^3+1. \]
Let us isolate \(x\). Subtracting \(1\) from both sides gives
\[ y-1=x^3, \]
and, taking cube roots,
\[ x=\sqrt[3]{y-1}. \]
Interchanging the names of the variables, the inverse function is
\[ f^{-1}(x)=\sqrt[3]{x-1}. \]
We may now apply the theorem on the continuity of the inverse function. The function \(f\) is continuous and strictly monotonic on the interval
\[ I=\mathbb{R}, \]
and hence its inverse is continuous on the image
\[ J=f(\mathbb{R})=\mathbb{R}. \]
Consequently,
\[ f^{-1}(x)=\sqrt[3]{x-1} \]
is continuous on the whole of \(\mathbb{R}\).
Problem 19 — level ★★★★☆
Determine the values of the parameters \(a,b\in\mathbb{R}\) for which the function
\[ f(x)= \begin{cases} x+1, & x<0,\\ ax+b, & 0\leq x\leq1,\\ 3x-1, & x>1 \end{cases} \]
is continuous on the whole of \(\mathbb{R}\).
Answer
The function is continuous on the whole of \(\mathbb{R}\) for
\[ a=1, \qquad b=1. \]
Solution
Each of the three expressions defining the function is polynomial, and is therefore continuous on its own interval.
The only points at which a problem of continuity could arise are consequently those at which the definition of the function changes:
\[ x=0 \qquad\text{and}\qquad x=1. \]
Let us begin with the point \(x=0\).
From the left, the function is
\[ f(x)=x+1, \]
so that
\[ \lim_{x\to0^-}f(x)=1. \]
At the point \(x=0\) itself, the middle expression applies:
\[ f(0)=a\cdot0+b=b. \]
The right-hand limit is likewise
\[ \lim_{x\to0^+}f(x)=b. \]
For continuity at \(0\) we must therefore have
\[ b=1. \]
Let us now turn to the point \(x=1\).
From the left, and at the point itself, the expression
\[ ax+b \]
holds. Thus
\[ f(1)=a+b \]
and
\[ \lim_{x\to1^-}f(x)=a+b. \]
From the right, on the other hand,
\[ f(x)=3x-1, \]
so that
\[ \lim_{x\to1^+}f(x)=3\cdot1-1=2. \]
For continuity at \(1\) we must have
\[ a+b=2. \]
We have already found
\[ b=1. \]
Substituting this into the preceding equation,
\[ a+1=2, \]
whence
\[ a=1. \]
Thus
\[ a=1, \qquad b=1. \]
With these values the function is continuous at the two junction points \(0\) and \(1\), while at every other point continuity follows immediately from the fact that each piece is polynomial.
The function is therefore continuous on the whole of \(\mathbb{R}\).
Problem 20 — level ★★★★★
Consider, on the interval \([-1,2]\), the function depending on the parameter \(a\):
\[ f(x)= \begin{cases} x^2+1, & -1\leq x\leq0,\\ ax+1, & 0<x\leq1,\\ 2x, & 1<x\leq2. \end{cases} \]
Determine the value of \(a\) for which \(f\) is continuous on \([-1,2]\). For this value of the parameter, determine the absolute minimum and maximum of the function, and show, by means of the intermediate value theorem, that there exists at least one point \(c\in(1,2)\) such that \(f(c)=3\).
Answer
The function is continuous on \([-1,2]\) for
\[ a=1. \]
For this value,
\[ \min_{x\in[-1,2]}f(x)=1, \]
attained at \(x=0\), while
\[ \max_{x\in[-1,2]}f(x)=4, \]
attained at \(x=2\).
Moreover, there exists at least one point \(c\in(1,2)\) such that \(f(c)=3\).
Solution
Each of the three pieces of the function is polynomial, and is therefore continuous within the interval on which it applies.
To establish the continuity of the whole function we must accordingly examine the two junction points:
\[ x=0 \qquad\text{and}\qquad x=1. \]
Let us consider first \(x=0\).
For \(x\leq0\),
\[ f(x)=x^2+1, \]
so that
\[ f(0)=0^2+1=1. \]
Moreover,
\[ \lim_{x\to0^-}f(x)=1. \]
From the right, the expression
\[ ax+1 \]
applies, so that
\[ \lim_{x\to0^+}f(x) = a\cdot0+1 = 1. \]
Hence
\[ \lim_{x\to0^-}f(x) = \lim_{x\to0^+}f(x) = f(0) = 1. \]
The function is thus continuous at \(x=0\) for every value of the parameter \(a\).
Let us turn to the point \(x=1\).
For \(0<x\leq1\),
\[ f(x)=ax+1, \]
so that
\[ f(1)=a+1 \]
and
\[ \lim_{x\to1^-}f(x)=a+1. \]
For \(x>1\),
\[ f(x)=2x, \]
so that
\[ \lim_{x\to1^+}f(x)=2. \]
For continuity at \(x=1\) we must impose
\[ a+1=2, \]
whence
\[ a=1. \]
With this value of the parameter the function becomes
\[ f(x)= \begin{cases} x^2+1, & -1\leq x\leq0,\\ x+1, & 0<x\leq1,\\ 2x, & 1<x\leq2. \end{cases} \]
We have verified that each of the three pieces is continuous and that the values match correctly at \(0\) and at \(1\). Moreover, at the endpoints \(-1\) and \(2\) the function has the appropriate one-sided continuity.
Hence \(f\) is continuous on the closed and bounded interval
\[ [-1,2]. \]
We may therefore apply Weierstrass's theorem: the function certainly attains an absolute minimum and an absolute maximum.
Let us examine the values taken on each of the three pieces separately.
On the first piece,
\[ -1\leq x\leq0, \]
we have
\[ f(x)=x^2+1. \]
On this interval
\[ 0\leq x^2\leq1, \]
so that
\[ 1\leq f(x)\leq2. \]
The minimum value on the first piece is
\[ f(0)=1, \]
while
\[ f(-1)=2. \]
On the second piece,
\[ 0<x\leq1, \]
we have
\[ f(x)=x+1. \]
Since
\[ 0<x\leq1, \]
it follows that
\[ 1<x+1\leq2. \]
On this piece, then, the values of the function exceed \(1\) and do not exceed \(2\).
On the third piece,
\[ 1<x\leq2, \]
we have
\[ f(x)=2x, \]
so that
\[ 2<2x\leq4, \]
the value \(4\) being actually attained at
\[ x=2. \]
Comparing the three pieces, the smallest value taken by the function as a whole is
\[ 1, \]
attained at \(x=0\). Hence
\[ \min_{x\in[-1,2]}f(x)=f(0)=1. \]
The largest value attained is instead
\[ 4, \]
attained at \(x=2\), so that
\[ \max_{x\in[-1,2]}f(x)=f(2)=4. \]
It remains to show that there exists at least one point \(c\in(1,2)\) such that
\[ f(c)=3. \]
Consider the function on the closed interval \([1,2]\). Since the function as a whole is continuous on \([-1,2]\), it is in particular continuous on \([1,2]\).
At the endpoints we have
\[ f(1)=2 \]
and
\[ f(2)=4. \]
The value \(3\) lies strictly between \(2\) and \(4\):
\[ f(1)=2<3<4=f(2). \]
By the intermediate value theorem there exists at least one point
\[ c\in(1,2) \]
such that
\[ f(c)=3. \]
In this instance it is also possible to identify such a point directly, since for \(1<x\leq2\) we have
\[ f(x)=2x. \]
Solving
\[ 2x=3 \]
gives
\[ x=\frac{3}{2}. \]
Indeed,
\[ 1<\frac{3}{2}<2 \]
and
\[ f\left(\frac{3}{2}\right) = 2\cdot\frac{3}{2} = 3. \]
The point guaranteed by the intermediate value theorem may thus be taken to be
\[ c=\frac{3}{2}. \]