Calculus

Applications of the Derivative

Applications of Derivative Logo

Global Extrema of Functions

  • Definition of Global Maximum and Global Minimum

    Consider a function \(y = f\left( x \right),\) which is supposed to be continuous on a closed interval \(\left[ {a,b} \right].\) If there exists a point \({x_0} \in \left[ {a,b} \right]\) such that \(f\left( x \right) \le f\left( {{x_0}} \right)\) for all \(x \in \left[ {a,b} \right],\) then we say that the function \(f\left( x \right)\) attains at \({x_0}\) the maximum (greatest) value over the interval \(\left[ {a,b} \right].\)

    The greatest value of the function \(f\left( x \right)\) on the interval \(\left[ {a,b} \right]\) is simultaneously the least upper bound of the range of the function on this interval and is denoted as

    \[
    {f\left( {{x_0}} \right) = \max\limits_{x \in \left[ {a,b} \right]} f\left( x \right) }
    = {\sup\limits_{x \in \left[ {a,b} \right]} f\left( x \right).}
    \]

    Similarly, if there exists a point \({x_0} \in \left[ {a,b} \right]\) such that \(f\left( x \right) \ge f\left( {{x_0}} \right)\) for all \(x \in \left[ {a,b} \right],\) then we say that the function \(f\left( x \right)\) attains at \({x_0}\) the minimum (least) value over the interval \(\left[ {a,b} \right].\)

    The least value of the function \(f\left( x \right)\) on the interval \(\left[ {a,b} \right]\) is simultaneously the greatest lower bound of the range of the function on this interval and can be written as

    \[
    {f\left( {{x_0}} \right) = \min\limits_{x \in \left[ {a,b} \right]} f\left( x \right) }
    = {\inf\limits_{x \in \left[ {a,b} \right]} f\left( x \right).}
    \]

    These concepts characterize the behavior of a function on a finite interval, in contrast to the local extremum, which describes the properties of the function in a small neighborhood of a point. Therefore, the maximum and minimum values of a function on an interval are often referred as the global (absolute) maximum or, respectively, the global minimum.

    The Weierstrass Extreme Value Theorem

    According to the Weierstrass extreme value theorem for continuous functions, if a function \(f\left( x \right)\) is continuous on a closed interval \(\left[ {a,b} \right],\) then it attains the least upper and greatest lower bounds on this interval.

    Karl Weierstrass
    Fig.1 Karl Weierstrass (1815-1897)

    The proof of this theorem is based on the Weierstrass boundedness theorem, which is stated as follows:

    If a function \(f\left( x \right)\) is continuous on the interval \(\left[ {a,b} \right],\) then it is bounded on it, i.e. there exists a real number \(M\) such that \(\left| {f\left( x \right)} \right| \le M\) for all \(x \in \left[ {a,b} \right].\)

    Returning to the extreme value theorem, we denote by \(M\) the least upper bound of the range of the function (or the maximum value of the function) on the interval \(\left[ {a,b} \right].\) Assume the contrary, that the least upper bound is not attained, i.e. assume that

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

    Consider the auxiliary function:

    \[\varphi \left( x \right) = \frac{1}{{M – f\left( x \right)}}.\]

    Since the denominator is not zero, then the function \(\varphi \left( x \right)\) is also continuous on \(\left[ {a,b} \right]\) and hence, by the boundedness theorem, it is bounded on this interval: \(\varphi \left( x \right) \le L,\) where \(L \gt 0.\) It follows from here that

    \[
    {\varphi \left( x \right) \le L,\;\;}\Rightarrow
    {\frac{1}{{M – f\left( x \right)}} \le L,\;\;}\Rightarrow
    {M – f\left( x \right) \ge \frac{1}{L},\;\;}\Rightarrow
    {f\left( x \right) \le M – \frac{1}{L}\;\forall \;x \in \left[ {a,b} \right].}
    \]

    In other words, the number \(M – {\large\frac{1}{L}\normalsize}\) will be the least upper bound of the function \(f\left( x \right)\), which contradicts the condition. (By the condition, the least upper bound of the function is equal to \(\left.{M}\right).\)

    In the same way we can prove that a function \(f\left( x \right)\) that is continuous in a closed interval \(\left[ {a,b} \right]\) attains in this interval its greatest lower bound (or the minimum value).

    Thus, according to the extreme value theorem, a function that is continuous on an interval always attains its maximum and minimum values on this interval.

    Finding the Maximum and Minimum Values of a Function

    Let a function \(y = f\left( x \right)\) be continuous on an interval \(\left[ {a,b} \right].\)

    If the function on this interval has local maxima at the points \({x_1},{x_2}, \ldots ,{x_n},\) then the maximum value of the function \(f\left( x \right)\) on the interval \(\left[ {a,b} \right]\) is equal to the greatest of the numbers

    \[{f\left( a \right),f\left( {{x_1}} \right),f\left( {{x_2}} \right), \ldots ,} \kern0pt{f\left( {{x_n}} \right),f\left( b \right).}\]

    Similarly, if the function on this interval has local minima at the points \({{\bar x}_1},{{\bar x}_2}, \ldots ,{{\bar x}_k},\) then the minimum value of the function \(f\left( x \right)\) on \(\left[ {a,b} \right]\) is equal to the least of the numbers

    \[{f\left( a \right),f\left( {{{\bar x}_1}} \right),f\left( {{{\bar x}_2}} \right), \ldots ,} \kern0pt{f\left( {{{\bar x}_k}} \right),f\left( b \right).}\]

    Thus, the global maximum (minimum) values of a function are attained either on the boundary of the interval (Figure \(2\)), or at the points of local extrema inside the interval (Figure \(3\)).

    The global maximum (minimum) of a function attained on the boundary of the interval
    Figure 2.
    The global maximum (minimum) values of a function coinciding with the points of local extrema
    Figure 3.

    Special Case \(1\)

    If there exists a unique extremum point \({x_1}\) inside the interval \(\left[ {a,b} \right]\) and this point is the local maximum (minimum), then the function attains the global maximum (minimum) at this point (Figure \(4\)).

    Special case 1 for global maximum (minimum) of a function
    Figure 4.

    Special Case \(2\)

    If the function \(y = f\left( x \right)\) has no critical points on the interval \(\left[ {a,b} \right],\) then the function has the global minimum at one endpoint of the interval and the global maximum on the other endpoint (Figure \(5\text{).}\)

    Special case 2 for global maximum (minimum) of a function
    Figure 5.

    Special Case \(3\)

    In practice, there are often cases when a differentiable and positively defined function \(f\left( x \right) \gt 0\) is given on an interval \(\left[ {a,b} \right]\) and is equal to zero at the endpoints of the interval: \(f\left( a \right) = f\left( b \right) = 0.\) If this function has a unique stationary point \({x_1}\) (where \(f’\left( {{x_1}} \right) = 0\)), then this point is not only the local maximum of the function, but also its global maximum on the interval \(\left[ {a,b} \right]\) (see Figure \(6\)).

    Special case 3 for global maximum (minimum) of a function
    Figure 6.

    In the examples below, find the global maximum and minimum of the function on the given interval.


  • Solved Problems

    Click a problem to see the solution.

    Example 1

    Find the global maximum and minimum of the function on the given interval: \[{f\left( x \right) = {x^2} – 2x + 5,\;\;\;}\kern-0.3pt{x \in \left[ { – 1,4} \right].}\]

    Example 2

    Calculate the difference \(d\) between the global maximum and global minimum values of \(f\left( x \right) = {x^2} – 4x + 6\) in the interval \(\left[ { – 2,4} \right].\)

    Example 3

    Find the global extrema of the function \(f\left( x \right) = {x^3} – 6{x^2} – 15x + 100\) on the interval \(\left[ { – 3,6} \right].\)

    Example 4

    Find the global maximum and minimum of the function on the given interval: \[{f\left( x \right) = x + \frac{2}{x},\;\;\;}\kern-0.3pt{x \in \left[ {0.5,2} \right].}\]

    Example 5

    Find the global maximum and global minimum of the function \(f\left( x \right) = {x^2} – \large{\frac{{16}}{x}}\normalsize\) on the interval \(\left[ { – 4, – 1} \right].\)

    Example 6

    Find the global extrema of the function \(f\left( x \right) = 3{x^4} – 6{x^2} + 2\) on the interval \(\left[ { – 2,2} \right].\)

    Example 7

    Find the global and global minimum of the function \(f\left( x \right) = {x^4} – 8{x^2} + 3\) within the interval \(\left[ { – 1,2} \right].\)

    Example 8

    Find the global extrema of the function \(f\left( x \right) = {x^4} – 2{x^2} – 1\) on \(\left[ {0, + \infty } \right).\)

    Example 9

    Find the global maximum and minimum of the function on the given interval: \[{f\left( x \right) = x\left| {x – 2} \right|,\;\;\;}\kern-0.3pt{x \in \left[ {0,3} \right].}\]

    Example 10

    Find the global maximum and minimum of the function \(f\left( x \right) = \sqrt {3 – 2x}\) on the interval \(\left[ { – 3,1} \right).\)

    Example 11

    Find the global extrema of \(f\left( x \right) = \sqrt {5 – 4x} \) on the interval \(\left[ { – 1,1} \right].\)

    Example 12

    Find the global maximum and global minimum of \(f\left( x \right) = {2^x}\) in the interval \(\left[ { – 1,3} \right].\)

    Example 13

    Find the global extrema of the function \(f\left( x \right) = \sqrt[3]{{{{\left( {x – 2} \right)}^2}}}\) on \(\left[ {1,4} \right].\)

    Example 14

    Find the global maximum and global minimum of \(f\left( x \right) = \frac{{{x^2}}}{{{2^x}}}\) in the interval \(\left[ { – 1,3} \right].\)

    Example 15

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

    Example 16

    Find the global extrema of \(f\left( x \right) = {x^2}{e^{ – x}}\) on the interval \(\left[ { – 1,1} \right].\)

    Example 17

    Find the global maximum and global minimum of \(f\left( x \right) = \sqrt[\large x\normalsize]{x}\) in the interval \(\left[ {2,3} \right].\)

    Example 18

    Find the global maximum and global minimum of the function \(f\left( x \right) = \sin x – x\) in the interval \(\left[ { – \pi ,\pi } \right].\)

    Example 19

    Find the global maximum and minimum of the function on the given interval: \[{f\left( x \right) = {\cos ^2}x – 2\sin x,\;\;\;}\kern-0.3pt{x \in \left[ {0,2\pi } \right].}\]

    Example 20

    Find the global maximum and minimum of the function on the given interval: \[{f\left( x \right) = \arctan \frac{{1 – x}}{{1 + x}},\;\;\;}\kern-0.3pt{x \in \left[ {0,1} \right].}\]

    Example 21

    Find the global maximum and minimum of the function on the given interval: \[{f\left( x \right) = \left| {\sin x – \frac{{\sqrt 3 }}{2}} \right|,\;\;\;}\kern-0.3pt{x \in \left[ {0,\pi } \right].}\]

    Example 1.

    Find the global maximum and minimum of the function on the given interval: \[{f\left( x \right) = {x^2} – 2x + 5,\;\;\;}\kern-0.3pt{x \in \left[ { – 1,4} \right].}\]

    Solution.

    The function is defined and differentiable for all \(x \in \mathbb{R}.\) Determine its stationary points:

    \[
    {f’\left( x \right) = 0,\;\;}\Rightarrow
    {{\left( {{x^2} – 2x + 5} \right)^\prime } = 0,\;\;}\Rightarrow
    {2x – 2 = 0,\;\;}\Rightarrow
    {x = 1.}
    \]

    This local extremum point belongs to the interval \(\left( { – 1,4} \right).\) We compute the values of the function at \(x = 1\) and at the endpoints of the interval:

    \[
    {f\left( 1 \right) = {1^2} – 2 \cdot 1 + 5 = 4,}\;\;\;\kern-0.3pt
    {f\left( { – 1} \right) = {\left( { – 1} \right)^2} – 2 \cdot \left( { – 1} \right) + 5 = 8,}\;\;\;\kern-0.3pt
    {f\left( 4 \right) = {4^2} – 2 \cdot 4 + 5 = 13.}
    \]

    Consequently, the maximum value of the function is equal \(f\left( 4 \right) = 13,\) and the minimum value is \(f\left( 1 \right) = 4.\)

    Example 2.

    Calculate the difference \(d\) between the global maximum and global minimum values of \(f\left( x \right) = {x^2} – 4x + 6\) in the interval \(\left[ { – 2,4} \right].\)

    Solution.

    Find the derivative:

    \[{f^\prime\left( x \right) = \left( {{x^2} – 4x + 6} \right)^\prime }={ 2x – 4.}\]

    Solve the equation \(f^\prime\left( c \right) = 0\) to determine the critical points:

    \[{f^\prime\left( c \right) = 0,}\;\; \Rightarrow {2c – 4 = 0,}\;\; \Rightarrow {c = 2.}\]

    We must evaluate \(f\left( x \right)\) at the critical point \(x = 2\) and at the endpoints \(x = -2,\) \(x = 4:\)

    \[{f\left( { – 2} \right) }={ {\left( { – 2} \right)^2} – 4 \cdot \left( { – 2} \right) + 6 }={ 18,}\]

    \[{f\left( 2 \right) }={ {2^2} – 4 \cdot 2 + 6 }={ 2,}\]

    \[{f\left( 4 \right) }={ {4^2} – 4 \cdot 4 + 6 }={ 6.}\]

    Thus, the global maximum value is 18 and the global minimum value is 2, so the difference \(d\) is equal to

    \[{d = {f_{\max }} – {f_{\min }} }={ 18 – 2 }={ 16.}\]

    Example 3.

    Find the global extrema of the function \(f\left( x \right) = {x^3} – 6{x^2} – 15x + 100\) on the interval \(\left[ { – 3,6} \right].\)

    Solution.

    global extrema of the cubic function f(x)=x^3-6x^2-15x+100
    Figure 7.

    Write the derivative of the function:

    \[{f^\prime\left( x \right) }={ \left( {{x^3} – 6{x^2} – 15x + 100} \right)^\prime }={ 3{x^2} – 12x – 15.}\]

    Determine the critical points by solving the equation \(f^\prime\left( c \right) = 0:\)

    \[{3{c^2} – 12c – 15 = 0,}\;\; \Rightarrow {3\left( {{c^2} – 4c – 5} \right) = 0,}\;\; \Rightarrow {3\left( {c + 1} \right)\left( {c – 5} \right) = 0,}\;\; \Rightarrow {{c_1} = – 1,\;}\kern0pt{{c_2} = 5.}\]

    Now we evaluate the function at these points and at the endpoints of the interval:

    \[{f\left( { – 3} \right) }={ {\left( { – 3} \right)^3} – 6 \cdot {\left( { – 3} \right)^2} -}\kern0pt{ 15 \cdot \left( { – 3} \right) + 100 }={ 64,}\]

    \[{f\left( { – 1} \right) }={ {\left( { – 1} \right)^3} – 6 \cdot {\left( { – 1} \right)^2} -}\kern0pt{ 15 \cdot \left( { – 1} \right) + 100 }={ 108,}\]

    \[{f\left( {5} \right) }={ {{5}^3} – 6 \cdot {{5} ^2} -}\kern0pt{ 15 \cdot {5} + 100 }={ 0,}\]

    \[{f\left( {6} \right) }={ {{6}^3} – 6 \cdot {{6} ^2} -}\kern0pt{ 15 \cdot {6} + 100 }={ 10.}\]

    We see that \(\left( {5,0} \right)\) is the global minimum point and \(\left( {-1,108} \right)\) is the global maximum point.

    Example 4.

    Find the global maximum and minimum of the function on the given interval: \[{f\left( x \right) = x + \frac{2}{x},\;\;\;}\kern-0.3pt{x \in \left[ {0.5,2} \right].}\]

    Solution.

    This function is not defined at \(x = 0,\) but this point is not included in the given interval. Differentiating, we find the points of local extremum:

    \[
    {f’\left( x \right) = {\left( {x + \frac{2}{x}} \right)^\prime } }
    = {1 – \frac{2}{{{x^2}}};}
    \]

    \[
    {f’\left( x \right) = 0,\;\;}\Rightarrow
    {1 – \frac{2}{{{x^2}}} = 0,\;\;}\Rightarrow
    {\frac{2}{{{x^2}}} = 1,\;\;}\Rightarrow
    {{x^2} = 2,\;\;}\Rightarrow
    {x = \pm \sqrt 2 .}
    \]

    As can be seen, only the point \(x = \sqrt 2\) falls in the interval \(\left[ {0.5,2} \right].\) Calculate the values of the function at the extremum point \(x = \sqrt 2\) and at the boundary points of the interval:

    \[
    {f\left( {\sqrt 2 } \right) = \sqrt 2 + \frac{2}{{\sqrt 2 }} = 2\sqrt 2 \approx 2.83;\;\;\;}\kern-0.3pt
    {f\left( {0.5} \right) = 0.5 + \frac{2}{{0.5}} = 4.5;\;\;\;}\kern-0.3pt
    {f\left( 2 \right) = 2 + \frac{2}{2} = 3.}
    \]

    So the maximum value of the function in this interval is equal to \(4.5\) at the point \(x = 0.5,\) and the minimum value is \(2.83\) at \(x = \sqrt 2.\)

    Example 5.

    Find the global maximum and global minimum of the function \(f\left( x \right) = {x^2} – \large{\frac{{16}}{x}}\normalsize\) on the interval \(\left[ { – 4, – 1} \right].\)

    Solution.

    Note that the domain of \(f\left( x \right)\) does not contain \(x = 0,\) and this point is not in the interval \(\left[ { – 4, – 1} \right].\)

    Compute the derivative:

    \[{f^\prime\left( x \right) = \left( {{x^2} – \frac{{16}}{x}} \right)^\prime }={ 2x + \frac{{16}}{{{x^2}}} }={ \frac{{2{x^3} + 16}}{{{x^2}}}.}\]

    Solve the equation \(f^\prime\left( c \right) = 0\) to determine the critical points:

    \[{f^\prime\left( c \right) = 0,}\;\; \Rightarrow {\frac{{2{c^3} + 16}}{{{c^2}}} = 0,}\;\; \Rightarrow {\left\{ {\begin{array}{*{20}{l}} {2{c^3} + 16 = 0}\\ {{c^2} \ne 0} \end{array}} \right.,}\;\; \Rightarrow {\left\{ {\begin{array}{*{20}{l}} {{c^3} = – 8}\\ {c \ne 0} \end{array}} \right.,}\;\; \Rightarrow {c = – 2.}\]

    Calculate the function values at the critical point and at the boundaries of the interval:

    \[{f\left( { – 4} \right) = {\left( { – 4} \right)^2} – \frac{{16}}{{\left( { – 4} \right)}} }={ 20,}\]

    \[{f\left( { – 2} \right) = {\left( { – 2} \right)^2} – \frac{{16}}{{\left( { – 2} \right)}} }={ 12,}\]

    \[{f\left( { – 1} \right) = {\left( { – 1} \right)^2} – \frac{{16}}{{\left( { – 1} \right)}} }={ 17.}\]

    Thus, the global (absolute) minimum occurs at \(\left( { – 2,12} \right),\) and the global maximum is attained at \(\left( { – 4,20} \right).\)

    Example 6.

    Find the global extrema of the function \(f\left( x \right) = 3{x^4} – 6{x^2} + 2\) on the interval \(\left[ { – 2,2} \right].\)

    Solution.

    This function is defined and differentiable on the whole real axis. In this case, all local extrema can be found from the equation \(f’\left( x \right) = 0:\)

    \[
    {f’\left( x \right) = {\left( {3{x^4} – 6{x^2} + 2} \right)^\prime } }
    = {12{x^3} – 12x }
    = {12x\left( {{x^2} – 1} \right) }
    = {12x\left( {x – 1} \right)\left( {x + 1} \right);}
    \]

    \[
    {f’\left( x \right) = 0,\;\;}\Rightarrow
    {12x\left( {x – 1} \right)\left( {x + 1} \right) = 0,\;\;}\Rightarrow
    {{x_1} = 0,\;{x_2} = – 1,\;{x_3} = 1.}
    \]

    As it can be seen, the function has three local extrema and all these points fall in the given interval \(\left[ { – 2,2} \right].\) Calculate the values of the function at the points of extremum and at the endpoints of the interval:

    \[
    {f\left( 0 \right) = 3 \cdot {0^4} – 6 \cdot {0^2} + 2 = 2;\;\;\;}\kern-0.3pt
    {f\left( { – 1} \right) = 3 \cdot {\left( { – 1} \right)^4} – 6 \cdot {\left( { – 1} \right)^2} + 2 = – 1;\;\;\;}\kern-0.3pt
    {f\left( { – 2} \right) = 3 \cdot {\left( { – 2} \right)^4} – 6 \cdot {\left( { – 2} \right)^2} + 2 = 26.}
    \]

    Since the function is even, we can write:

    \[
    {f\left( 1 \right) = f\left( { – 1} \right) = – 1,}\;\;\;\kern-0.3pt
    {f\left( 2 \right) = f\left( { – 2} \right) = 26.}
    \]

    Thus, the function has the minimum value \(-1\) at two points: at \(x = -1\) and \(x = 1.\) The maximum value \(26\) is also attained at two points: at \(x = -2\) and \(x = 2.\) A schematic graph of the function is given in Figure \(8.\)

    Global maximum and minimum points of the quartic function y=3x^4-6x^2+2
    Figure 8.
    Page 1
    Problems 1-6
    Page 2
    Problems 7-21