Calculus

Applications of the Derivative

Applications of Derivative Logo

Increasing and Decreasing Functions

  • Definition of an Increasing and Decreasing Function

    Let \(y = f\left( x \right)\) be a differentiable function on an interval \(\left( {a,b} \right).\) If for any two points \({x_1},{x_2} \in \left( {a,b} \right)\) such that \({x_1} < {x_2},\) there holds the inequality \(f\left( {{x_1}} \right) \le f\left( {{x_2}} \right),\) the function is called increasing (or non-decreasing) in this interval.

    Definition of increasing function
    Figure 1.

    If this inequality is strict, i.e. \(f\left( {{x_1}} \right) \lt f\left( {{x_2}} \right),\) then the function \(y = f\left( x \right)\) is said to be strictly increasing on the interval \(\left( {a,b} \right).\)

    Similarly, we define a decreasing (or non-increasing) and a strictly decreasing function.

    Definition of decreasing function
    Figure 2.

    These concepts can be formulated in a more compact form. A function \(y = f\left( x \right)\) is called

    • increasing (non-decreasing) on the interval \(\left( {a,b} \right)\) if
      \[ {\forall\;{x_1},{x_2} \in \left( {a,b} \right):\;}\kern-0.3pt {{x_1} \lt {x_2}} \Rightarrow {f\left( {{x_1}} \right) \le f\left( {{x_2}} \right);} \]
    • strictly increasing on the interval \(\left( {a,b} \right)\) if
      \[ {\forall\;{x_1},{x_2} \in \left( {a,b} \right):\;}\kern-0.3pt {{x_1} \lt {x_2}} \Rightarrow {f\left( {{x_1}} \right) \lt f\left( {{x_2}} \right);} \]
    • decreasing (non-increasing) on the interval \(\left( {a,b} \right)\) if
      \[ {\forall\;{x_1},{x_2} \in \left( {a,b} \right):\;}\kern-0.3pt {{x_1} \lt {x_2}} \Rightarrow {f\left( {{x_1}} \right) \ge f\left( {{x_2}} \right);} \]
    • strictly decreasing on the interval \(\left( {a,b} \right)\) if
      \[ {\forall\;{x_1},{x_2} \in \left( {a,b} \right):\;}\kern-0.3pt {{x_1} \lt {x_2}} \Rightarrow {f\left( {{x_1}} \right) \gt f\left( {{x_2}} \right).} \]

    It is clear that a non-decreasing function can contain strictly increasing intervals and intervals where the function is constant. This is schematically illustrated in Figures \(3-6.\)

    increasing function
    Figure 3.
    strictly increasing function
    Figure 4.
    decreasing function
    Figure 5.
    strictly decreasing function
    Figure 6.

    If a function \(f\left( x \right)\) is differentiable on the interval \(\left( {a,b} \right)\) and belongs to one of the four considered types (i.e. it is increasing, strictly increasing, decreasing, or strictly decreasing), this function is called monotonic on this interval.

    The concept of increasing and decreasing functions can also be defined for a single point \({x_0}.\) In this case, we consider a small \(\delta\)-neighborhood \(\left( {{x_0} – \delta ,{x_0} + \delta } \right)\) of this point. A function \(y = f\left( x \right)\) is strictly increasing at \({x_0}\) if there exists a number \(\delta \gt 0\) such that

    \[{\forall\;x \in \left( {{x_0} – \delta ,{x_0}} \right)} \Rightarrow {f\left( x \right) \lt f\left( {{x_0}} \right);}\]

    \[{\forall\;x \in \left( {{x_0}, {x_0} + \delta} \right)} \Rightarrow {f\left( x \right) \gt f\left( {{x_0}} \right).}\]

    Similarly, we can define a function \(y = f\left( x \right)\), which is strictly decreasing at the point \({x_0}.\)

    Criteria for Increasing and Decreasing Functions

    Again consider a function \(y = f\left( x \right)\) assuming it is differentiable on an interval \(\left( {a,b} \right).\) To determine if the function is increasing or decreasing on the interval, we use the sign of the first derivative of the function.

    Theorem \(1.\)

    In order for the function \(y = f\left( x \right)\) to be increasing on the interval \(\left( {a,b} \right),\) it is necessary and sufficient that the first derivative of the function be non-negative everywhere in this interval:

    \[f’\left( x \right) \ge 0\;\forall\;x \in \left( {a,b} \right).\]

    A similar criterion applies to the case of a function that is decreasing on the interval \(\left( {a,b} \right):\)

    \[f’\left( x \right) \le 0\;\forall\;x \in \left( {a,b} \right).\]

    We prove both (necessary and sufficient) parts of the theorem for the case of an increasing function.

    Necessary condition.

    Consider an arbitrary point \({x_0} \in \left( {a,b} \right).\) If the function \(y = f\left( x \right)\) is increasing on \(\left( {a,b} \right),\) then by definition, we can write:

    \[{\forall\;x \in \left( {a,b} \right):x \gt {x_0}} \Rightarrow {f\left( x \right) \gt f\left( {{x_0}} \right);}\]

    \[{\forall\;x \in \left( {a,b} \right):x \lt {x_0}} \Rightarrow {f\left( x \right) \lt f\left( {{x_0}} \right).}\]

    It can be seen that in both cases the following inequality holds:

    \[
    {\frac{{f\left( x \right) – f\left( {{x_0}} \right)}}{{x – {x_0}}} \ge 0,}\;\;\;\kern-0.3pt
    {\text{where}\;\;x \ne {x_0}.}
    \]

    In the limit as \(x \to {x_0}\), the left-hand side of the inequality is equal to the derivative of the function at the point \({x_0},\) that is by the limit sign preservation property:

    \[
    {\lim\limits_{x \to {x_0}} \frac{{f\left( x \right) – f\left( {{x_0}} \right)}}{{x – {x_0}}} }
    = {f’\left( {{x_0}} \right) \ge 0.}
    \]

    This relationship is valid for any \({x_0} \in \left( {a,b} \right).\)

    Consider the sufficient condition, that is the converse statement. Suppose that the derivative \(f’\left( x \right)\) of a function \(y = f\left( x \right)\) is non-negative in the interval \(\left( {a,b} \right):\)

    \[f’\left( {{x_0}} \right) \ge 0\;\forall\; x \in \left( {a,b} \right).\]

    If \({x_1}\) and \({x_2}\) are two arbitrary points of the interval such that \({x_1}< {x_2},\) then by Lagrange’s theorem we can write:

    \[{f\left( {{x_2}} \right) – f\left( {{x_1}} \right)} = {f’\left( c \right)\left( {{x_2} – {x_1}} \right),}\]

    where \(c \in \left[ {{x_1},{x_2}} \right],\) \(\Rightarrow c \in \left( {a,b} \right).\)

    Since \(f’\left( c \right) \ge 0,\) then the right-hand side of the equality is non-negative. Consequently,

    \[f\left( {{x_2}} \right) \ge f\left( {{x_1}} \right).\]

    i.e. the function \(y = f\left( x \right)\) is increasing in the interval \(\left( {a,b} \right).\)

    Consider now the cases of a strictly increasing and strictly decreasing function. There exists a similar theorem that describes the necessary and sufficient conditions. Omitting the proof, we state it for the case of a strictly increasing function.

    Theorem \(2.\)

    Suppose that a function \(y = f\left( x \right)\) is differentiable on an interval \(\left( {a,b} \right).\) In order for the function to be strictly increasing in this interval, it is necessary and sufficient that the following conditions are satisfied:

    1. \(f’\left( x \right) \ge 0\;\forall\;x \in \left( {a,b} \right);\)
    2. \(f’\left( x \right)\) is not identically equal to zero at any interval \(\left[ {{x_1},{x_2}} \right] \in \left( {a,b} \right).\)

    The condition \(1\) is contained in Theorem \(1\) and is an indication of a non-decreasing function. The additional condition \(2\) is required in order to exclude the intervals of constancy, in which the derivative of \(f\left( x \right)\) is identically zero.

    In practice (when finding the intervals of monotonicity), the sufficient condition for a strictly increasing or a strictly decreasing function is commonly used. Theorem \(2\) implies the following wording of the sufficient criterion:

    If the condition \(f’\left( x \right) \gt 0\) is satisfied for all \(x \in \left( {a,b} \right)\), except perhaps only a few distinct points where \(f’\left( x \right) = 0,\) then the function \(f\left( x \right)\) is strictly increasing in this interval.

    Accordingly, the condition \(f’\left( x \right) \lt 0\) defines a strictly decreasing function.

    The number of points where \(f’\left( x \right) = 0\) is usually finite. According to Theorem \(2,\) they can not tightly fill any subinterval of the interval \(\left( {a,b} \right).\)

    We also give a criterion for increasing/decreasing functions at a point:

    Theorem \(3.\)

    Let \({x_0} \in \left( {a,b} \right).\)

    • If \(f’\left( {{x_0}} \right) \gt 0\), then the function \(f\left( x \right)\) is strictly increasing at the point \({x_0};\)
    • If \(f’\left( {{x_0}} \right) \lt 0\), then the function \(f\left( x \right)\) is strictly decreasing at the point \({x_0}.\)

    Properties of Monotonic Functions

    Increasing and decreasing functions have certain algebraic properties, which may be useful in the investigation of functions. Here are some of them:

    1. If the functions \(f\) and \(g\) are increasing (decreasing) on the interval \(\left( {a,b} \right),\) then the sum of the functions \(f + g\) is also increasing (decreasing) on this interval.
    2. If the function \(f\) is increasing (decreasing) on the interval \(\left( {a,b} \right),\) then the opposite function \(-f\) is decreasing (increasing) on this interval.
    3. If the function \(f\) is increasing (decreasing) on the interval \(\left( {a,b} \right),\) then the inverse function \(\large\frac{1}{f}\normalsize\) is decreasing (increasing) on this interval.
    4. If the functions \(f\) and \(g\) are increasing (decreasing) on the interval \(\left( {a,b} \right)\) and moreover, \(f \ge 0\), \(g \ge 0\), then the product of the functions \(fg\) is also increasing (decreasing) on this interval.
    5. If the function \(g\) is increasing (decreasing) on the interval \(\left( {a,b} \right)\) and the function \(f\) is increasing (decreasing) on \(\left( {c,d} \right)\) where \(g:\left( {a,b} \right) \to \left( {c,d} \right),\) then the composition of functions \(f \circ g\) (i.e. the composite function \(y = f\left( {g\left( x \right)} \right)\) is also increasing (decreasing) on the interval \(\left( {a,b} \right).\)

  • Solved Problems

    Click a problem to see the solution.

    Example 1

    Indicate the intervals where the function \(y = f\left( x \right)\) is strictly increasing (Figure \(7\)).

    Example 2

    Indicate the intervals where the function \(y = f\left( x \right)\) is decreasing (Figure \(8\)).

    Example 3

    Using the definition of monotonicity prove that the function \(f\left( x \right) = {x^2} + 1\) is strictly increasing for \(x \ge 0.\)

    Example 4

    Using the definition of monotonicity prove that the cubic function \(f\left( x \right) = {x^3}\) is strictly increasing for all \(x \in \mathbb{R}.\)

    Example 5

    Using the properties of monotonic functions prove that the function \(f\left( x \right) = {x^4} + 3{x^2}\) is strictly increasing for \(x \ge 0.\)

    Example 6

    Using the definition of monotonicity prove that the function \(f\left( x \right) = \cos x\) is strictly decreasing on the interval \(\left[ {0,\pi } \right].\)

    Example 7

    Show that the function \(f\left( x \right) = {x^3} – 3{x^2} + 6x – 1\) is strictly increasing on \(\mathbb{R}.\)

    Example 8

    For what values of \(x\) is the function \(f\left( x \right) = {x^4} – 2{x^2}\) strictly increasing?

    Example 9

    What is the length \(L\) of the interval on which the function \(f\left( x \right) = {x^3} – 6{x^2} – 15x + 8\) is decreasing?

    Example 10

    What is the length \(L\) of the interval on which the function \(f\left( x \right) = {x^4}{e^{ – x}}\) is increasing?

    Example 11

    Find the intervals of monotonicity of the function \(f\left( x \right) = {x^3} – 12x + 5.\)

    Example 12

    Find the intervals of monotonicity of the function \(f\left( x \right) = x + \sin x.\)

    Example 13

    Find the intervals of monotonicity of the function \[f\left( x \right) = \frac{x}{{{x^2} + 1}}.\]

    Example 14

    Find the intervals on which the function \({y = {x^x}\,}\kern0pt{\left( {x \gt 0} \right)}\) is increasing and decreasing.

    Example 15

    Find the intervals on which the function \(f\left( x \right) = x – \large{\frac{4}{{{x^2}}}}\normalsize\) is increasing and decreasing.

    Example 16

    Find the intervals of monotonicity of the function
    \[{f\left( x \right) = \frac{{\sqrt x }}{{x – 1}}\;\;}\kern-0.3pt{\left( {x \ge 0} \right).}\]

    Example 17

    Find the intervals of monotonicity of the function \(f\left( x \right) = x\ln x.\)

    Example 18

    Find the intervals of monotonicity of the function \(f\left( x \right) = {x^2}{e^{ – x}}.\)

    Example 19

    Find the intervals of monotonicity of the function \(f\left( x \right) = \sqrt {x – {x^2}}.\)

    Example 20

    Find the intervals on which the function \(f\left( x \right) = {x^3} + \large{\frac{1}{{{x^3}}}}\normalsize\) is strictly decreasing.

    Example 21

    Find the intervals of monotonicity of the function \(f\left( x \right) = {\large\frac{{1 – \sin x}}{{\cos x}}\normalsize}.\)

    Example 22

    Find the intervals on which the function \(f\left( x \right) = \large{\frac{x}{{\ln x}}}\normalsize\) is increasing and decreasing.

    Example 23

    Find the intervals on which the function \(f\left( x \right) = \large{\frac{{{x^2} + 1}}{x}}\normalsize\) is increasing and decreasing.

    Example 24

    Find all values of the parameter \(a,\) for which the function \(f\left( x \right) = {x^3} – 6{x^2} + ax\) is strictly increasing in the whole domain.

    Example 25

    Find all values of the parameter \(a,\) for which the equation \({x^3} – 6{x^2} + 9x + a = 0\) has three distinct real roots.

    Example 26

    Determine the number of roots of the cubic equation \({x^3} – 12x + a = 0\) depending on the parameter \(a.\)

    Example 1.

    Indicate the intervals where the function \(y = f\left( x \right)\) is strictly increasing (Figure \(7\)).

    Solution.

    function with increasing and decreasing intervals
    Figure 7.

    A function is strictly increasing when the \(y-\)value increases as the \(x-\)value increases. One can see that the given function is strictly increasing on the intervals \(\left( { – 5, – 1} \right)\) and \(\left( {3,6} \right).\)

    Example 2.

    Indicate the intervals where the function \(y = f\left( x \right)\) is decreasing (Figure \(8\)).

    Solution.

    example of increasing and decreasing function
    Figure 8.

    According to the definition, a function is decreasing on an interval if \(f\left( {{x_1}} \right) \ge f\left( {{x_2}} \right)\) for any two points \({x_1} \le {x_2}.\)

    Thus, a decreasing interval may also contain points where the function has a constant value. (This is not true for a strictly decreasing function.)

    In our case, we see that the function is decreasing on the interval \(\left( { – 2,6} \right).\)

    Example 3.

    Using the definition of monotonicity prove that the function \(f\left( x \right) = {x^2} + 1\) is strictly increasing for \(x \ge 0.\)

    Solution.

    We take two arbitrary points \({x_1}\) and \({x_2}\) such that

    \[0 \le {x_1} \lt {x_2}.\]

    Consider the difference between the values of the function at these points:

    \[
    {f\left( {{x_2}} \right) – f\left( {{x_1}} \right) }
    = {\left( {x_2^2 + 1} \right) – \left( {x_1^2 + 1} \right) }
    = {x_2^2 – x_1^2 }
    = {\left( {{x_2} – {x_1}} \right)\left( {{x_2} + {x_1}} \right).}
    \]

    It is obvious that in the last expression \({{x_2} – {x_1}} \gt 0\) and \({{x_2} + {x_1}} \gt 0\) (since, by assumption, only non-negative values of \(x\) are considered). As a result, we have

    \[
    {\left( {{x_2} – {x_1}} \right)\left( {{x_2} + {x_1}} \right) > 0,\;\;}\Rightarrow
    {f\left( {{x_2}} \right) – f\left( {{x_1}} \right) > 0.}
    \]

    This means by definition that the function \(f\left( x \right) = {x^2} + 1\) is strictly increasing on the given interval.

    Example 4.

    Using the definition of monotonicity prove that the cubic function \(f\left( x \right) = {x^3}\) is strictly increasing for all \(x \in \mathbb{R}.\)

    Solution.

    We choose two arbitrary points \({{x_1}}\) and \({{x_2}}\) such that \({x_1} \lt {x_2}.\) Consider the difference:

    \[f\left( {{x_2}} \right) – f\left( {{x_1}} \right) = x_2^3 – x_1^3.\]

    Factoring it as the difference of cubes, we obtain:

    \[
    {x_2^3 – x_1^3 }
    = {\left( {{x_2} – {x_1}} \right)\left( {x_2^2 + {x_1}{x_2} + x_1^2} \right).}
    \]

    In the second bracket we can get a perfect square:

    \[
    {x_2^2 + {x_1}{x_2} + x_1^2 }
    = {x_1^2 + 2 \cdot {x_1} \cdot \frac{{{x_2}}}{2} + \frac{{x_2^2}}{4} + \frac{{3x_2^2}}{4} }
    = {{\left( {{x_1} + \frac{{{x_2}}}{2}} \right)^2} + \frac{{3x_2^2}}{4} \gt 0.}
    \]

    Hence it is clear that the quadratic expression is always positive (it is equal to zero only if \({x_1} = {x_2} = 0,\) which contradicts the condition \({x_1} \lt {x_2}.\))

    Thus, \(f\left( {{x_2}} \right) – f\left( {{x_1}} \right) \gt 0,\) if \({x_2} – {x_1} \gt 0,\) i.e. the function \(f\left( x \right) = {x^3}\) is strictly increasing.

    Example 5.

    Using the properties of monotonic functions prove that the function \(f\left( x \right) = {x^4} + 3{x^2}\) is strictly increasing for \(x \ge 0.\)

    Solution.

    This function is the sum of the functions \({x^4}\) and \(3{x^2}.\)

    The first function \({x^4}\) can be considered as the product of two identical functions \({x^2}\). Example \(1\) shows that the quadratic function \({x^2}\) is strictly increasing for \(x \ge 0.\) Hence, the function \({x^4}\) is also strictly increasing for \(x \ge 0\) by the property \(4.\)

    The second term \(3{x^2}\) is the triple sum of the functions \({x^2}\), so this term is also strictly increasing according to the property \(1.\)

    Hence, the original function \(f\left( x \right) = {x^4} + 3{x^2}\) is the sum of two strictly increasing functions and consequently is also strictly increasing when \(x \ge 0.\)

    Example 6.

    Using the definition of monotonicity prove that the function \(f\left( x \right) = \cos x\) is strictly decreasing on the interval \(\left[ {0,\pi } \right].\)

    Solution.

    Let the points \({x_1}\), \({x_2}\) lie in the given interval \(\left[ {0,\pi } \right]\) and the following condition is satisfied: \({x_1} \lt {x_2}.\) Consider the difference:

    \[f\left( {{x_2}} \right) – f\left( {{x_1}} \right) = \cos {x_2} – \cos {x_1}.\]

    Convert it by the cosine difference identity:

    \[
    {\cos {x_2} – \cos {x_1} }
    = { – 2\sin \frac{{{x_2} + {x_1}}}{2}\sin \frac{{{x_2} – {x_1}}}{2}.}
    \]

    Since \({x_1},{x_2} \in \left[ {0,\pi } \right],\) then the half sum is bounded by the double inequality

    \[0 \lt \frac{{{x_2} + {x_1}}}{2} \lt \pi .\]

    Similarly, the half difference (assuming that \({x_2} > {x_1}\)) satisfies the inequality

    \[0 \lt \frac{{{x_2} – {x_1}}}{2} \lt \frac{\pi }{2}.\]

    For these ranges of angles, the sine is always positive. Therefore

    \[
    {\cos {x_2} – \cos {x_1} }
    = { – 2\sin \frac{{{x_2} + {x_1}}}{2}\sin \frac{{{x_2} – {x_1}}}{2} \lt 0.}
    \]

    Thus, the following relationship holds:

    \[{x_2} \gt {x_1} \Rightarrow f\left( {{x_2}} \right) \lt f\left( {{x_1}} \right),\]

    that is the cosine function is strictly decreasing on the interval \(\left[ {0,\pi } \right].\)

    Example 7.

    Show that the function \(f\left( x \right) = {x^3} – 3{x^2} + 6x – 1\) is strictly increasing on \(\mathbb{R}.\)

    Solution.

    Find the derivative:

    \[{{f^\prime\left( x \right)} = \left( {{x^3} – 3{x^2} + 6x – 1} \right)^\prime }={{3}{x^2} – 6x + 6.}\]

    Notice that the discriminant of the quadratic function is negative:

    \[{D = {b^2} – 4ac }={ {\left( { – 6} \right)^2} – 4 \cdot 3 \cdot 6 }={ 36 – 72 }={ – 36 \lt 0.}\]

    Therefore, the quadratic function has no zeros and has the same sign over the interval \(\left( { – \infty ,\infty } \right).\)

    We choose \(x = 0\) to evaluate the sign of the derivative:

    \[{f^\prime\left( 0 \right) = 3 \cdot {0^2} – 6 \cdot 0 + 6 }=\cssId{element16}{ 6 \gt 0.}\]

    Hence, the function is strictly increasing on \(\mathbb{R}.\)

    Example 8.

    For what values of \(x\) is the function \(f\left( x \right) = {x^4} – 2{x^2}\) strictly increasing?

    Solution.

    Sign chart for the first derivative of f(x)=x^4-2x^2
    Figure 9.

    Calculate the derivative:

    \[{f^\prime\left( x \right) = \left( {{x^4} – 2{x^2}} \right)^\prime }={ 4{x^3} – 4x }={ 4x\left( {{x^2} – 1} \right) }={ 4x\left( {x – 1} \right)\left( {x + 1} \right).}\]

    The derivative is zero at the points \({x_1} = – 1,\) \({x_1} = 0,\) \({x_3} = 1.\)

    Using the interval method we find the intervals where the derivative has a constant sign (see the sign chart above).

    Hence, the function is increasing on \(\left( { – 1,0} \right)\) and \(\left( {1, + \infty } \right).\)

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    Problems 1-8
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    Problems 9-26