This collection presents 20 practice problems on increasing functions, decreasing functions, monotonic functions, and strictly monotonic functions. Each exercise is solved step by step, with particular attention to comparing the values the function takes at two points \(x_1\) and \(x_2\) of the domain.
To decide whether a function is increasing or decreasing on a set \(X\), it is not enough to look at a few points of the graph: one must check a property that holds for every pair of points \(x_1,x_2\in X\) with \(x_1<x_2\).
Recall that a function \(f:X\to\mathbb R\) is increasing on \(X\) if
\[ x_1<x_2 \implies f(x_1)\le f(x_2), \]
whereas it is decreasing on \(X\) if
\[ x_1<x_2 \implies f(x_1)\ge f(x_2). \]
If the inequalities are strict, the function is instead said to be strictly increasing or strictly decreasing. The exercises that follow show how to apply these definitions, how to recognize functions that are monotonic on intervals, and how to tell ordinary monotonicity apart from strict monotonicity.
Exercise 1 — level ★☆☆☆☆
Determine whether the function
\[ f:\mathbb R\to\mathbb R,\qquad f(x)=2x+1 \]
is increasing, decreasing, strictly increasing, or strictly decreasing on \(\mathbb R\).
Answer
The function is strictly increasing on \(\mathbb R\). Consequently, it is also increasing on \(\mathbb R\).
Solution
To study monotonicity directly from the definition, take any two points \(x_1,x_2\in\mathbb R\) such that
\[ x_1<x_2. \]
We must compare \(f(x_1)\) and \(f(x_2)\). Since
\[ f(x)=2x+1, \]
we have
\[ f(x_1)=2x_1+1 \]
and
\[ f(x_2)=2x_2+1. \]
Multiplying both sides of \(x_1<x_2\) by \(2\), which is positive, we obtain
\[ 2x_1<2x_2. \]
Adding \(1\) to both sides yields
\[ 2x_1+1<2x_2+1, \]
that is,
\[ f(x_1)<f(x_2). \]
We have thus shown that, for all \(x_1,x_2\in\mathbb R\),
\[ x_1<x_2 \implies f(x_1)<f(x_2). \]
By definition, the function is strictly increasing on \(\mathbb R\).
Since every strictly increasing function is also increasing, we conclude that \(f\) is increasing on \(\mathbb R\) as well.
Exercise 2 — level ★☆☆☆☆
Determine whether the function
\[ f:\mathbb R\to\mathbb R,\qquad f(x)=-3x+4 \]
is increasing, decreasing, strictly increasing, or strictly decreasing on \(\mathbb R\).
Answer
The function is strictly decreasing on \(\mathbb R\). Consequently, it is also decreasing on \(\mathbb R\).
Solution
Take any two points \(x_1,x_2\in\mathbb R\) such that
\[ x_1<x_2. \]
We must compare the values \(f(x_1)\) and \(f(x_2)\). Since
\[ f(x)=-3x+4, \]
we have
\[ f(x_1)=-3x_1+4 \]
and
\[ f(x_2)=-3x_2+4. \]
From the inequality
\[ x_1<x_2 \]
multiplying both sides by \(-3\), a negative number, reverses the inequality:
\[ -3x_1>-3x_2. \]
Adding \(4\) to both sides, we obtain
\[ -3x_1+4>-3x_2+4. \]
that is,
\[ f(x_1)>f(x_2). \]
We have thus shown that, for all \(x_1,x_2\in\mathbb R\),
\[ x_1<x_2 \implies f(x_1)>f(x_2). \]
By definition, \(f\) is strictly decreasing on \(\mathbb R\).
Since every strictly decreasing function is also decreasing, the function is also decreasing on \(\mathbb R\).
Exercise 3 — level ★☆☆☆☆
Determine whether the constant function
\[ f:\mathbb R\to\mathbb R,\qquad f(x)=5 \]
is increasing, decreasing, strictly increasing, or strictly decreasing on \(\mathbb R\).
Answer
The function is both increasing and decreasing on \(\mathbb R\), but it is neither strictly increasing nor strictly decreasing.
Solution
Consider any two points \(x_1,x_2\in\mathbb R\) such that
\[ x_1<x_2. \]
Since the function is constant, its value is always equal to \(5\). Thus
\[ f(x_1)=5 \]
and
\[ f(x_2)=5. \]
In particular,
\[ f(x_1)=f(x_2). \]
From this equality there follow both
\[ f(x_1)\le f(x_2) \]
and
\[ f(x_1)\ge f(x_2). \]
Hence, for all \(x_1,x_2\in\mathbb R\) with \(x_1<x_2\), both conditions hold:
\[ f(x_1)\le f(x_2) \]
and
\[ f(x_1)\ge f(x_2). \]
By definition, the function is therefore both increasing and decreasing on \(\mathbb R\).
However, it is not strictly increasing. Indeed, to be strictly increasing it would have to satisfy
\[ f(x_1)<f(x_2) \]
for every \(x_1<x_2\), but here the two values are always equal.
Likewise, it is not strictly decreasing, because
\[ f(x_1)>f(x_2) \]
does not hold for every \(x_1<x_2\).
We conclude that the constant function is increasing and decreasing, but neither strictly increasing nor strictly decreasing.
Exercise 4 — level ★★☆☆☆
Determine whether the function
\[ f:\mathbb R\to\mathbb R,\qquad f(x)=x^2 \]
is increasing, decreasing, or non-monotonic on \(\mathbb R\).
Answer
The function is not monotonic on \(\mathbb R\): it is neither increasing nor decreasing on its whole domain.
Solution
To decide whether \(f(x)=x^2\) is increasing on \(\mathbb R\), we would have to verify that, for all \(x_1,x_2\in\mathbb R\),
\[ x_1<x_2 \implies f(x_1)\le f(x_2). \]
But it is enough to exhibit a single pair of points violating this condition to show that the function is not increasing.
Choose
\[ x_1=-1,\qquad x_2=0. \]
Clearly
\[ -1<0, \]
that is, \(x_1<x_2\). However,
\[ f(-1)=(-1)^2=1 \]
whereas
\[ f(0)=0^2=0. \]
Hence
\[ f(-1)>f(0). \]
We have found two points \(x_1<x_2\) with \(f(x_1)>f(x_2)\). This contradicts the definition of an increasing function. Therefore \(f\) is not increasing on \(\mathbb R\).
Let us now check whether the function is decreasing on \(\mathbb R\). To be decreasing it would have to satisfy, for all \(x_1,x_2\in\mathbb R\),
\[ x_1<x_2 \implies f(x_1)\ge f(x_2). \]
Again, it is enough to find a pair that violates the condition.
Choose
\[ x_1=0,\qquad x_2=1. \]
We have
\[ 0<1, \]
but
\[ f(0)=0 \]
and
\[ f(1)=1. \]
Hence
\[ f(0)<f(1). \]
We have found two points \(x_1<x_2\) with \(f(x_1)<f(x_2)\). This contradicts the definition of a decreasing function. Therefore \(f\) is not decreasing on \(\mathbb R\).
Since the function is neither increasing nor decreasing on \(\mathbb R\), we conclude that \(f(x)=x^2\) is not monotonic on \(\mathbb R\).
Exercise 5 — level ★★☆☆☆
Determine whether the function
\[ f:[0,+\infty)\to\mathbb R,\qquad f(x)=x^2 \]
is increasing, decreasing, strictly increasing, or strictly decreasing on \([0,+\infty)\).
Answer
The function is strictly increasing on \([0,+\infty)\). Consequently, it is also increasing on \([0,+\infty)\).
Solution
To study monotonicity on \([0,+\infty)\), take any two points
\[ x_1,x_2\in[0,+\infty) \]
such that
\[ x_1<x_2. \]
Since \(x_1\) and \(x_2\) belong to \([0,+\infty)\), they are both nonnegative. In particular,
\[ 0\le x_1<x_2. \]
We want to compare \(f(x_1)\) and \(f(x_2)\). Since
\[ f(x)=x^2, \]
we have
\[ f(x_1)=x_1^2 \]
and
\[ f(x_2)=x_2^2. \]
From the inequality
\[ 0\le x_1<x_2 \]
it follows that
\[ x_1^2<x_2^2. \]
Indeed, we may write
\[ x_2^2-x_1^2=(x_2-x_1)(x_2+x_1). \]
Since \(x_2>x_1\), we have
\[ x_2-x_1>0. \]
Moreover, since \(x_1\ge 0\) and \(x_2>x_1\), we also have
\[ x_2+x_1>0. \]
Hence
\[ (x_2-x_1)(x_2+x_1)>0, \]
that is,
\[ x_2^2-x_1^2>0. \]
From this it follows that
\[ x_1^2<x_2^2. \]
Therefore
\[ f(x_1)<f(x_2). \]
We have shown that, for all \(x_1,x_2\in[0,+\infty)\),
\[ x_1<x_2 \implies f(x_1)<f(x_2). \]
By definition, \(f\) is strictly increasing on \([0,+\infty)\).
Consequently, \(f\) is also increasing on \([0,+\infty)\).
Exercise 6 — level ★★☆☆☆
Determine whether the function
\[ f:(-\infty,0]\to\mathbb R,\qquad f(x)=x^2 \]
is increasing, decreasing, strictly increasing, or strictly decreasing on \((-\infty,0]\).
Answer
The function is strictly decreasing on \((-\infty,0]\). Consequently, it is also decreasing on \((-\infty,0]\).
Solution
Take any two points
\[ x_1,x_2\in(-\infty,0] \]
such that
\[ x_1<x_2. \]
Since both points belong to \((-\infty,0]\), we have
\[ x_1<x_2\le 0. \]
We must compare
\[ f(x_1)=x_1^2 \]
and
\[ f(x_2)=x_2^2. \]
Consider the difference
\[ x_1^2-x_2^2. \]
Factor it as a difference of squares:
\[ x_1^2-x_2^2=(x_1-x_2)(x_1+x_2). \]
Since \(x_1<x_2\), we have
\[ x_1-x_2<0. \]
Moreover, since \(x_1<x_2\le 0\), both numbers are nonpositive and \(x_1\) at least is strictly negative. Hence
\[ x_1+x_2<0. \]
The product of two negative numbers is positive, so
\[ (x_1-x_2)(x_1+x_2)>0. \]
Therefore
\[ x_1^2-x_2^2>0. \]
From this inequality it follows that
\[ x_1^2>x_2^2. \]
that is,
\[ f(x_1)>f(x_2). \]
We have thus shown that, for all \(x_1,x_2\in(-\infty,0]\),
\[ x_1<x_2 \implies f(x_1)>f(x_2). \]
By definition, \(f\) is strictly decreasing on \((-\infty,0]\).
Consequently, \(f\) is also decreasing on \((-\infty,0]\).
Exercise 7 — level ★★☆☆☆
Determine whether the function
\[ f:\mathbb R\setminus\{0\}\to\mathbb R,\qquad f(x)=\frac{1}{x} \]
is decreasing on its whole domain.
Answer
The function is not decreasing on its whole domain \(\mathbb R\setminus\{0\}\).
Solution
The domain of the function is
\[ \mathbb R\setminus\{0\}=(-\infty,0)\cup(0,+\infty). \]
To be decreasing on its whole domain, the function would have to satisfy the following condition: for all \(x_1,x_2\in\mathbb R\setminus\{0\}\),
\[ x_1<x_2 \implies f(x_1)\ge f(x_2). \]
To show that the function is not decreasing on its whole domain, it suffices to find a pair of points in the domain that violates this condition.
Choose
\[ x_1=-1,\qquad x_2=1. \]
Both belong to the domain, since they are different from \(0\), and
\[ -1<1. \]
We compute the values of the function:
\[ f(-1)=\frac{1}{-1}=-1 \]
and
\[ f(1)=\frac{1}{1}=1. \]
Hence
\[ f(-1)<f(1). \]
But a decreasing function would have to satisfy
\[ f(-1)\ge f(1), \]
since \(-1<1\).
The pair \(x_1=-1\), \(x_2=1\) therefore violates the definition of a decreasing function.
Hence the function
\[ f(x)=\frac{1}{x} \]
is not decreasing on its whole domain \(\mathbb R\setminus\{0\}\).
This does not contradict the fact that \(f\) is strictly decreasing separately on \((-\infty,0)\) and on \((0,+\infty)\). Monotonicity must always be understood relative to the set on which it is studied.
Exercise 8 — level ★★☆☆☆
Determine whether the function
\[ f:(0,+\infty)\to\mathbb R,\qquad f(x)=\frac{1}{x} \]
is increasing, decreasing, strictly increasing, or strictly decreasing on \((0,+\infty)\).
Answer
The function is strictly decreasing on \((0,+\infty)\). Consequently, it is also decreasing on \((0,+\infty)\).
Solution
Take any two points
\[ x_1,x_2\in(0,+\infty) \]
such that
\[ x_1<x_2. \]
Since \(x_1\) and \(x_2\) belong to \((0,+\infty)\), they are both positive:
\[ 0<x_1<x_2. \]
We want to compare
\[ f(x_1)=\frac{1}{x_1} \]
and
\[ f(x_2)=\frac{1}{x_2}. \]
Since
\[ 0<x_1<x_2, \]
dividing \(1\) by a larger positive number yields a smaller value. Algebraically, we compare the two fractions:
\[ \frac{1}{x_1}-\frac{1}{x_2} = \frac{x_2-x_1}{x_1x_2}. \]
The numerator is positive, since
\[ x_2-x_1>0. \]
The denominator is positive as well, since \(x_1>0\) and \(x_2>0\). Hence
\[ x_1x_2>0. \]
It follows that
\[ \frac{x_2-x_1}{x_1x_2}>0. \]
Thus
\[ \frac{1}{x_1}-\frac{1}{x_2}>0, \]
that is,
\[ \frac{1}{x_1}>\frac{1}{x_2}. \]
Therefore
\[ f(x_1)>f(x_2). \]
We have shown that, for all \(x_1,x_2\in(0,+\infty)\),
\[ x_1<x_2 \implies f(x_1)>f(x_2). \]
By definition, \(f\) is strictly decreasing on \((0,+\infty)\).
Consequently, \(f\) is also decreasing on \((0,+\infty)\).
Exercise 9 — level ★★☆☆☆
Determine whether the function
\[ f:(-\infty,0)\to\mathbb R,\qquad f(x)=\frac{1}{x} \]
is increasing, decreasing, strictly increasing, or strictly decreasing on \((-\infty,0)\).
Answer
The function is strictly decreasing on \((-\infty,0)\). Consequently, it is also decreasing on \((-\infty,0)\).
Solution
Take any two points
\[ x_1,x_2\in(-\infty,0) \]
such that
\[ x_1<x_2. \]
Since \(x_1\) and \(x_2\) belong to \((-\infty,0)\), they are both negative. Hence
\[ x_1<x_2<0. \]
We want to compare
\[ f(x_1)=\frac{1}{x_1} \]
and
\[ f(x_2)=\frac{1}{x_2}. \]
Consider the difference:
\[ \frac{1}{x_1}-\frac{1}{x_2} = \frac{x_2-x_1}{x_1x_2}. \]
Since \(x_1<x_2\), we have
\[ x_2-x_1>0. \]
Moreover, \(x_1\) and \(x_2\) are both negative, so their product is positive:
\[ x_1x_2>0. \]
Therefore
\[ \frac{x_2-x_1}{x_1x_2}>0. \]
Thus
\[ \frac{1}{x_1}-\frac{1}{x_2}>0, \]
that is,
\[ \frac{1}{x_1}>\frac{1}{x_2}. \]
We have therefore obtained
\[ f(x_1)>f(x_2). \]
This holds for every pair \(x_1,x_2\in(-\infty,0)\) with \(x_1<x_2\). By definition, the function is strictly decreasing on \((-\infty,0)\).
Consequently, \(f\) is also decreasing on \((-\infty,0)\).
Exercise 10 — level ★★★☆☆
Determine whether the function
\[ f:\mathbb R\to\mathbb R,\qquad f(x)=x^3 \]
is increasing, decreasing, strictly increasing, or strictly decreasing on \(\mathbb R\).
Answer
The function is strictly increasing on \(\mathbb R\). Consequently, it is also increasing on \(\mathbb R\).
Solution
We argue directly from the definition. Take any two points \(x_1,x_2\in\mathbb R\) such that
\[ x_1<x_2. \]
We must show that
\[ f(x_1)<f(x_2), \]
that is,
\[ x_1^3<x_2^3. \]
Consider the difference
\[ x_2^3-x_1^3. \]
Factor the difference of cubes:
\[ x_2^3-x_1^3=(x_2-x_1)(x_2^2+x_1x_2+x_1^2). \]
Since \(x_1<x_2\), we have
\[ x_2-x_1>0. \]
It remains to observe that
\[ x_2^2+x_1x_2+x_1^2>0. \]
Indeed, we may write
\[ x_2^2+x_1x_2+x_1^2 = \left(x_2+\frac{x_1}{2}\right)^2+\frac{3}{4}x_1^2. \]
This quantity is always nonnegative and, in our case, cannot vanish while \(x_1<x_2\). Hence it is positive.
Therefore the product
\[ (x_2-x_1)(x_2^2+x_1x_2+x_1^2) \]
is positive. Hence
\[ x_2^3-x_1^3>0. \]
From this it follows that
\[ x_1^3<x_2^3. \]
that is,
\[ f(x_1)<f(x_2). \]
We have shown that, for all \(x_1,x_2\in\mathbb R\),
\[ x_1<x_2 \implies f(x_1)<f(x_2). \]
By definition, \(f(x)=x^3\) is strictly increasing on \(\mathbb R\).
This example is important because it shows that a function can be proved strictly increasing directly from the definition, by comparing the values it takes at two arbitrary points of the domain.
Exercise 11 — level ★★★☆☆
Determine whether the function
\[ f:\mathbb R\to\mathbb R,\qquad f(x)=x^2-4x+1 \]
is increasing, decreasing, or non-monotonic on \(\mathbb R\).
Answer
The function is not monotonic on \(\mathbb R\). It is strictly decreasing on \((-\infty,2]\) and strictly increasing on \([2,+\infty)\).
Solution
We rewrite the function by completing the square:
\[ f(x)=x^2-4x+1=(x-2)^2-3. \]
This form shows that the value of the function depends on the squared distance of \(x\) from the number \(2\).
We first study the function on the interval \([2,+\infty)\). Take any two points
\[ x_1,x_2\in[2,+\infty) \]
such that
\[ x_1<x_2. \]
Since \(x_1\ge 2\) and \(x_2\ge 2\), we have
\[ 0\le x_1-2<x_2-2. \]
Squaring, since both sides are nonnegative, we obtain
\[ (x_1-2)^2<(x_2-2)^2. \]
Subtracting \(3\) from both sides:
\[ (x_1-2)^2-3<(x_2-2)^2-3. \]
that is,
\[ f(x_1)<f(x_2). \]
Hence \(f\) is strictly increasing on \([2,+\infty)\).
We now study the function on the interval \((-\infty,2]\). Take any two points
\[ x_1,x_2\in(-\infty,2] \]
such that
\[ x_1<x_2. \]
Then
\[ x_1-2<x_2-2\le 0. \]
Multiplying by \(-1\) reverses the inequality:
\[ 2-x_1>2-x_2\ge 0. \]
Since both sides are nonnegative, squaring gives
\[ (2-x_1)^2>(2-x_2)^2. \]
But
\[ (2-x)^2=(x-2)^2. \]
Hence
\[ (x_1-2)^2>(x_2-2)^2. \]
Subtracting \(3\) from both sides:
\[ (x_1-2)^2-3>(x_2-2)^2-3. \]
that is,
\[ f(x_1)>f(x_2). \]
Hence \(f\) is strictly decreasing on \((-\infty,2]\).
Finally, the function is not monotonic on all of \(\mathbb R\), because it first decreases and then increases. We can also verify this with two counterexamples.
Indeed,
\[ f(1)=1-4+1=-2 \]
whereas
\[ f(2)=4-8+1=-3. \]
Since \(1<2\) but \(f(1)>f(2)\), the function is not increasing on \(\mathbb R\).
Moreover,
\[ f(2)=-3 \]
and
\[ f(3)=9-12+1=-2. \]
Since \(2<3\) but \(f(2)<f(3)\), the function is not decreasing on \(\mathbb R\).
We conclude that \(f\) is not monotonic on \(\mathbb R\), but it is strictly decreasing on \((-\infty,2]\) and strictly increasing on \([2,+\infty)\).
Exercise 12 — level ★★★☆☆
Determine whether the function
\[ f:\mathbb R\to\mathbb R,\qquad f(x)=-x^2+6x-5 \]
is increasing, decreasing, or non-monotonic on \(\mathbb R\).
Answer
The function is not monotonic on \(\mathbb R\). It is strictly increasing on \((-\infty,3]\) and strictly decreasing on \([3,+\infty)\).
Solution
We rewrite the function by completing the square:
\[ f(x)=-x^2+6x-5=-(x-3)^2+4. \]
This form shows that the function attains its maximum value at \(x=3\), since the term \((x-3)^2\) is always nonnegative.
We first study the function on \((-\infty,3]\). Take any two points
\[ x_1,x_2\in(-\infty,3] \]
such that
\[ x_1<x_2. \]
Then
\[ x_1-3<x_2-3\le 0. \]
Multiplying by \(-1\), we obtain
\[ 3-x_1>3-x_2\ge 0. \]
Since both sides are nonnegative, squaring gives
\[ (3-x_1)^2>(3-x_2)^2. \]
Since
\[ (3-x)^2=(x-3)^2, \]
we obtain
\[ (x_1-3)^2>(x_2-3)^2. \]
Multiplying by \(-1\) reverses the inequality:
\[ -(x_1-3)^2<-(x_2-3)^2. \]
Adding \(4\) to both sides:
\[ -(x_1-3)^2+4<-(x_2-3)^2+4. \]
that is,
\[ f(x_1)<f(x_2). \]
Hence \(f\) is strictly increasing on \((-\infty,3]\).
We now study the function on \([3,+\infty)\). Take any two points
\[ x_1,x_2\in[3,+\infty) \]
such that
\[ x_1<x_2. \]
Then
\[ 0\le x_1-3<x_2-3. \]
Squaring:
\[ (x_1-3)^2<(x_2-3)^2. \]
Multiplying by \(-1\), we obtain
\[ -(x_1-3)^2>-(x_2-3)^2. \]
Adding \(4\):
\[ -(x_1-3)^2+4>-(x_2-3)^2+4. \]
that is,
\[ f(x_1)>f(x_2). \]
Hence \(f\) is strictly decreasing on \([3,+\infty)\).
The function is not monotonic on all of \(\mathbb R\), because it increases up to \(x=3\) and then decreases.
Indeed, choosing \(x_1=2\) and \(x_2=3\), we have \(2<3\), but
\[ f(2)=-4+12-5=3 \]
and
\[ f(3)=-9+18-5=4. \]
Hence \(f(2)<f(3)\), which rules out that the function is decreasing on \(\mathbb R\).
Moreover, choosing \(x_1=3\) and \(x_2=4\), we have \(3<4\), but
\[ f(3)=4 \]
and
\[ f(4)=-16+24-5=3. \]
Hence \(f(3)>f(4)\), which rules out that the function is increasing on \(\mathbb R\).
We conclude that \(f\) is not monotonic on \(\mathbb R\), but it is strictly increasing on \((-\infty,3]\) and strictly decreasing on \([3,+\infty)\).
Exercise 13 — level ★★★☆☆
Determine whether the function
\[ f:\mathbb R\to\mathbb R,\qquad f(x)=|x| \]
is increasing, decreasing, or non-monotonic on \(\mathbb R\). Then determine the intervals on which it is monotonic.
Answer
The function is not monotonic on \(\mathbb R\). It is strictly decreasing on \((-\infty,0]\) and strictly increasing on \([0,+\infty)\).
Solution
The absolute value function is defined by
\[ |x|= \begin{cases} -x & \text{if } x<0,\\ x & \text{if } x\ge 0. \end{cases} \]
We first study the function on \([0,+\infty)\). If \(x\ge 0\), then
\[ |x|=x. \]
So take any two points
\[ x_1,x_2\in[0,+\infty) \]
such that
\[ x_1<x_2. \]
Since \(f(x)=x\) on \([0,+\infty)\), we obtain
\[ f(x_1)=x_1 \]
and
\[ f(x_2)=x_2. \]
From the inequality \(x_1<x_2\) it follows directly that
\[ f(x_1)<f(x_2). \]
Hence \(f\) is strictly increasing on \([0,+\infty)\).
We now study the function on \((-\infty,0]\). If \(x\le 0\), then
\[ |x|=-x. \]
Take any two points
\[ x_1,x_2\in(-\infty,0] \]
such that
\[ x_1<x_2. \]
Since \(f(x)=-x\) on \((-\infty,0]\), we obtain
\[ f(x_1)=-x_1 \]
and
\[ f(x_2)=-x_2. \]
From the inequality
\[ x_1<x_2 \]
multiplying by \(-1\) reverses the inequality:
\[ -x_1>-x_2. \]
Hence
\[ f(x_1)>f(x_2). \]
Hence \(f\) is strictly decreasing on \((-\infty,0]\).
The function is not monotonic on all of \(\mathbb R\). Indeed, choosing
\[ x_1=-1,\qquad x_2=0, \]
we have \(x_1<x_2\), but
\[ f(-1)=1>0=f(0). \]
This rules out that \(f\) is increasing on \(\mathbb R\).
Moreover, choosing
\[ x_1=0,\qquad x_2=1, \]
we have \(x_1<x_2\), but
\[ f(0)=0<1=f(1). \]
This rules out that \(f\) is decreasing on \(\mathbb R\).
Hence \(f(x)=|x|\) is not monotonic on \(\mathbb R\), but it is strictly decreasing on \((-\infty,0]\) and strictly increasing on \([0,+\infty)\).
Exercise 14 — level ★★★☆☆
Determine whether the function
\[ f:\mathbb R\to\mathbb R,\qquad f(x)=-|x| \]
is increasing, decreasing, or non-monotonic on \(\mathbb R\). Then determine the intervals on which it is monotonic.
Answer
The function is not monotonic on \(\mathbb R\). It is strictly increasing on \((-\infty,0]\) and strictly decreasing on \([0,+\infty)\).
Solution
The function is
\[ f(x)=-|x|. \]
Since
\[ |x|= \begin{cases} -x & \text{if } x<0,\\ x & \text{if } x\ge 0, \end{cases} \]
we obtain
\[ -|x|= \begin{cases} x & \text{if } x<0,\\ -x & \text{if } x\ge 0. \end{cases} \]
We first study the function on \((-\infty,0]\). On this interval the function behaves as
\[ f(x)=x. \]
If \(x_1,x_2\in(-\infty,0]\) and
\[ x_1<x_2, \]
then
\[ f(x_1)=x_1 \]
and
\[ f(x_2)=x_2. \]
Thus from the inequality \(x_1<x_2\) it follows that
\[ f(x_1)<f(x_2). \]
Hence \(f\) is strictly increasing on \((-\infty,0]\).
We now study the function on \([0,+\infty)\). On this interval the function behaves as
\[ f(x)=-x. \]
If \(x_1,x_2\in[0,+\infty)\) and
\[ x_1<x_2, \]
then, multiplying by \(-1\), we obtain
\[ -x_1>-x_2, \]
that is,
\[ f(x_1)>f(x_2). \]
Hence \(f\) is strictly decreasing on \([0,+\infty)\).
The function is not monotonic on all of \(\mathbb R\). Indeed, it increases up to \(x=0\) and then decreases.
To see that it is not increasing on all of \(\mathbb R\), choose
\[ x_1=0,\qquad x_2=1. \]
We have \(0<1\), but
\[ f(0)=0>-1=f(1). \]
This contradicts the function being increasing.
To see that it is not decreasing on all of \(\mathbb R\), choose
\[ x_1=-1,\qquad x_2=0. \]
We have \(-1<0\), but
\[ f(-1)=-1<0=f(0). \]
This contradicts the function being decreasing.
Hence \(f(x)=-|x|\) is not monotonic on \(\mathbb R\), but it is strictly increasing on \((-\infty,0]\) and strictly decreasing on \([0,+\infty)\).
Exercise 15 — level ★★★★☆
Determine whether the function
\[ f:\mathbb R\to\mathbb R,\qquad f(x)= \begin{cases} x & \text{if } x<0,\\ 0 & \text{if } x\ge 0 \end{cases} \]
is increasing, decreasing, strictly increasing, or strictly decreasing on \(\mathbb R\).
Answer
The function is increasing on \(\mathbb R\), but it is not strictly increasing.
Solution
The function is defined piecewise:
\[ f(x)= \begin{cases} x & \text{if } x<0,\\ 0 & \text{if } x\ge 0. \end{cases} \]
We must verify that, for all \(x_1,x_2\in\mathbb R\) with
\[ x_1<x_2, \]
we have
\[ f(x_1)\le f(x_2). \]
We consider the possible cases.
First case: \(x_1<x_2<0\). Here both points are negative, so
\[ f(x_1)=x_1,\qquad f(x_2)=x_2. \]
Since \(x_1<x_2\), it follows that
\[ f(x_1)<f(x_2), \]
and so, in particular,
\[ f(x_1)\le f(x_2). \]
Second case: \(x_1<0\le x_2\). Here
\[ f(x_1)=x_1 \]
and
\[ f(x_2)=0. \]
Since \(x_1<0\), we have
\[ f(x_1)=x_1<0=f(x_2). \]
So in this case too \(f(x_1)\le f(x_2)\).
Third case: \(0\le x_1<x_2\). Here both points are nonnegative, so
\[ f(x_1)=0,\qquad f(x_2)=0. \]
Consequently
\[ f(x_1)=f(x_2), \]
and therefore
\[ f(x_1)\le f(x_2). \]
In all possible cases we have obtained
\[ x_1<x_2 \implies f(x_1)\le f(x_2). \]
By definition, the function is increasing on \(\mathbb R\).
The function is not, however, strictly increasing. Indeed, choosing
\[ x_1=1,\qquad x_2=2, \]
we have \(x_1<x_2\), but
\[ f(1)=0=f(2). \]
So \(f(x_1)<f(x_2)\) does not hold for every pair \(x_1<x_2\). Hence the function is increasing, but not strictly increasing.
Exercise 16 — level ★★★★☆
Determine whether the function
\[ f:\mathbb R\to\mathbb R,\qquad f(x)= \begin{cases} 1 & \text{if } x<0,\\ -x & \text{if } x\ge 0 \end{cases} \]
is increasing, decreasing, strictly increasing, or strictly decreasing on \(\mathbb R\).
Answer
The function is decreasing on \(\mathbb R\), but it is not strictly decreasing.
Solution
The function is defined piecewise:
\[ f(x)= \begin{cases} 1 & \text{if } x<0,\\ -x & \text{if } x\ge 0. \end{cases} \]
To show that \(f\) is decreasing on \(\mathbb R\), we must verify that, for all \(x_1,x_2\in\mathbb R\),
\[ x_1<x_2 \implies f(x_1)\ge f(x_2). \]
We consider all possible cases.
First case: \(x_1<x_2<0\).
Here both points are negative. So, by the definition of the function,
\[ f(x_1)=1 \qquad \text{and} \qquad f(x_2)=1. \]
Therefore
\[ f(x_1)=f(x_2), \]
and so, in particular,
\[ f(x_1)\ge f(x_2). \]
Second case: \(x_1<0\le x_2\).
Here \(x_1\) is negative, whereas \(x_2\) is nonnegative. So
\[ f(x_1)=1 \]
and
\[ f(x_2)=-x_2. \]
Since \(x_2\ge 0\), we have
\[ -x_2\le 0. \]
Hence
\[ f(x_2)\le 0. \]
But
\[ f(x_1)=1, \]
so certainly
\[ f(x_1)\ge f(x_2). \]
Third case: \(0\le x_1<x_2\).
Here both points are nonnegative. So the function is given by
\[ f(x)=-x. \]
Therefore
\[ f(x_1)=-x_1 \qquad \text{and} \qquad f(x_2)=-x_2. \]
From the inequality
\[ x_1<x_2 \]
multiplying by \(-1\) reverses the inequality:
\[ -x_1>-x_2, \]
that is,
\[ f(x_1)>f(x_2). \]
In particular, in this case too we have
\[ f(x_1)\ge f(x_2). \]
In all possible cases we have shown that, if \(x_1<x_2\), then
\[ f(x_1)\ge f(x_2). \]
By definition, \(f\) is therefore decreasing on \(\mathbb R\).
The function is not, however, strictly decreasing. Indeed, choosing
\[ x_1=-2,\qquad x_2=-1, \]
we have
\[ -2<-1, \]
but, since both points are negative,
\[ f(-2)=1 \qquad \text{and} \qquad f(-1)=1. \]
Hence
\[ f(-2)=f(-1). \]
To be strictly decreasing it would instead have to satisfy
\[ f(-2)>f(-1). \]
This condition does not hold. Hence the function is decreasing on \(\mathbb R\), but not strictly decreasing.
Exercise 17 — level ★★★★☆
Determine whether the function
\[ f:\mathbb R\to\mathbb R,\qquad f(x)= \begin{cases} x & \text{if } x\le 0,\\ x+1 & \text{if } x>0 \end{cases} \]
is increasing, decreasing, strictly increasing, or strictly decreasing on \(\mathbb R\).
Answer
The function is strictly increasing on \(\mathbb R\). Consequently, it is also increasing on \(\mathbb R\).
Solution
The function is defined piecewise:
\[ f(x)= \begin{cases} x & \text{if } x\le 0,\\ x+1 & \text{if } x>0. \end{cases} \]
To show that \(f\) is strictly increasing on \(\mathbb R\), we must verify that, for all \(x_1,x_2\in\mathbb R\),
\[ x_1<x_2 \implies f(x_1)<f(x_2). \]
We consider all possible cases.
First case: \(x_1<x_2\le 0\).
Here both points belong to the first piece of the function. So
\[ f(x_1)=x_1 \]
and
\[ f(x_2)=x_2. \]
Since \(x_1<x_2\), we obtain directly
\[ f(x_1)<f(x_2). \]
Second case: \(0<x_1<x_2\).
Here both points belong to the second piece of the function. So
\[ f(x_1)=x_1+1 \]
and
\[ f(x_2)=x_2+1. \]
Adding \(1\) to both sides of \(x_1<x_2\), it follows that
\[ x_1+1<x_2+1, \]
that is,
\[ f(x_1)<f(x_2). \]
Third case: \(x_1\le 0<x_2\).
Here \(x_1\) belongs to the first piece, whereas \(x_2\) belongs to the second. So
\[ f(x_1)=x_1 \]
and
\[ f(x_2)=x_2+1. \]
Since \(x_1\le 0\) and \(x_2>0\), we have
\[ x_1\le 0<x_2<x_2+1. \]
In particular,
\[ x_1<x_2+1, \]
that is,
\[ f(x_1)<f(x_2). \]
In all possible cases we have shown that
\[ x_1<x_2 \implies f(x_1)<f(x_2). \]
By definition, the function is strictly increasing on \(\mathbb R\).
Consequently, it is also increasing on \(\mathbb R\).
Exercise 18 — level ★★★★☆
Determine whether the function
\[ f:\mathbb R\to\mathbb R,\qquad f(x)= \begin{cases} x & \text{if } x<0,\\ x-1 & \text{if } x\ge 0 \end{cases} \]
is increasing, decreasing, or non-monotonic on \(\mathbb R\).
Answer
The function is non-monotonic on \(\mathbb R\): it is neither increasing nor decreasing.
Solution
The function is defined piecewise:
\[ f(x)= \begin{cases} x & \text{if } x<0,\\ x-1 & \text{if } x\ge 0. \end{cases} \]
Observe that the function increases on each of the two pieces taken separately. Indeed, for \(x<0\) we have \(f(x)=x\), whereas for \(x\ge 0\) we have \(f(x)=x-1\).
This, however, is not enough to conclude that the function is increasing on all of \(\mathbb R\). One must also check what happens at the transition from negative to nonnegative values.
To show that \(f\) is not increasing on \(\mathbb R\), we find two points \(x_1,x_2\in\mathbb R\) such that
\[ x_1<x_2 \]
but
\[ f(x_1)>f(x_2). \]
Choose
\[ x_1=-\frac12,\qquad x_2=0. \]
We have
\[ -\frac12<0. \]
We compute the values of the function:
\[ f\left(-\frac12\right)=-\frac12, \]
since \(-\frac12<0\), whereas
\[ f(0)=0-1=-1, \]
since \(0\ge 0\).
Hence
\[ f\left(-\frac12\right)=-\frac12>-1=f(0). \]
We have found two points \(x_1<x_2\) with \(f(x_1)>f(x_2)\). This contradicts the definition of an increasing function. Therefore \(f\) is not increasing on \(\mathbb R\).
To show that \(f\) is not decreasing on \(\mathbb R\), we find two points \(x_1,x_2\in\mathbb R\) such that
\[ x_1<x_2 \]
but
\[ f(x_1)<f(x_2). \]
Choose
\[ x_1=1,\qquad x_2=2. \]
We have
\[ 1<2. \]
Since both points are nonnegative, we obtain
\[ f(1)=1-1=0 \]
and
\[ f(2)=2-1=1. \]
Hence
\[ f(1)<f(2). \]
This pair contradicts the definition of a decreasing function. Therefore \(f\) is not decreasing on \(\mathbb R\).
Since the function is neither increasing nor decreasing on \(\mathbb R\), we conclude that it is non-monotonic on \(\mathbb R\).
Exercise 19 — level ★★★★☆
Prove that every strictly increasing function is injective.
Answer
Every strictly increasing function is injective.
Solution
Let
\[ f:X\to\mathbb R \]
be a strictly increasing function on a set \(X\subseteq\mathbb R\).
We want to prove that \(f\) is injective. By definition, we must show that distinct elements of the domain have distinct images.
So take any two points
\[ x_1,x_2\in X \]
such that
\[ x_1\ne x_2. \]
Since \(x_1\) and \(x_2\) are two distinct real numbers, exactly one of the following two possibilities holds:
\[ x_1<x_2 \]
or
\[ x_2<x_1. \]
If \(x_1<x_2\), then, since \(f\) is strictly increasing, we obtain
\[ f(x_1)<f(x_2). \]
In particular,
\[ f(x_1)\ne f(x_2). \]
If instead \(x_2<x_1\), then, again because \(f\) is strictly increasing, we obtain
\[ f(x_2)<f(x_1). \]
In this case too it follows that
\[ f(x_1)\ne f(x_2). \]
In either case, from \(x_1\ne x_2\) it follows that
\[ f(x_1)\ne f(x_2). \]
By definition, \(f\) is injective.
We therefore conclude that every strictly increasing function is injective.
Exercise 20 — level ★★★★★
Determine whether the following statement is true or false:
\[ \text{If a function is injective, then it is monotonic.} \]
If it is false, provide a counterexample.
Answer
The statement is false. There exist injective functions that are not monotonic.
Solution
The statement to be examined is:
\[ \text{if a function is injective, then it is monotonic.} \]
This statement is false. To prove it, it suffices to construct an injective function that is neither increasing nor decreasing.
Consider the function
\[ f:\{1,2,3\}\to\mathbb R \]
defined by
\[ f(1)=1,\qquad f(2)=3,\qquad f(3)=2. \]
First we verify that \(f\) is injective. The values taken by the function are
\[ 1,\qquad 3,\qquad 2. \]
These three values are all distinct. Therefore distinct elements of the domain have distinct images. By definition, \(f\) is injective.
We now verify that \(f\) is not increasing.
If \(f\) were increasing, then for every pair \(x_1,x_2\in\{1,2,3\}\) with \(x_1<x_2\), we would have
\[ f(x_1)\le f(x_2). \]
But
\[ 2<3, \]
while
\[ f(2)=3 \]
and
\[ f(3)=2. \]
Hence
\[ f(2)>f(3). \]
This contradicts the definition of an increasing function. Therefore \(f\) is not increasing.
We now verify that \(f\) is not decreasing.
If \(f\) were decreasing, then for every pair \(x_1,x_2\in\{1,2,3\}\) with \(x_1<x_2\), we would have
\[ f(x_1)\ge f(x_2). \]
But
\[ 1<2, \]
while
\[ f(1)=1 \]
and
\[ f(2)=3. \]
Hence
\[ f(1)<f(2). \]
This contradicts the definition of a decreasing function. Therefore \(f\) is not decreasing.
The function is injective, but it is neither increasing nor decreasing. Hence it is not monotonic.
This counterexample shows that injectivity does not imply monotonicity.