1.3 Maximierungs- und Minimierungsprobleme
Aus Online Mathematik Brückenkurs 2
Theorie | Übungen |
Inhalt:
- Kurven zeichnen
- Maximierungs- und Minimierungsprobleme
Lernziele:
Nach diesem Abschnitt sollten Sie folgendes können:
- The definition of strictly increasing, strictly decreasing, local maximum, local minimum, global maximum, global minimum.
- That if \displaystyle f^{\,\prime}>0 in an interval then \displaystyle f is strictly increasing in the interval, and that if \displaystyle f^{\,\prime}<0 in an interval then \displaystyle f is strictly decreasing in the interval.
- To locate stationary points and, by studying the sign of the derivative, classify them as local maxima, local minima, and stationary points of inflexion.
- To sketch the graph of a function by constructing a table of signs for the derivative.
- To determine global and local maxima and minima by 1) studying the sign of the derivative, 2) considering points where Die Funktion is not differentiable, 3) examining the endpoints of the interval where Die Funktion is defined.
- To use the sign of the second derivative to distinguish between local maxima and local minima.
Steigende und fallende Funktionen
Man sagt dass eine Funktion steigend ist falls ihre Ableitung positiv ist, und fallend falls ihre Ableitung negativ ist.
Die formelle Definitionen lauten:
Eine Funktion f ist steigend in einen bestimmten Intervall, falls für alle \displaystyle x_1 und \displaystyle x_2 im Intervall:
\displaystyle x_1 < x_2\quad\Rightarrow\quad f(x_1) \le f(x_2)\,\mbox{.} |
Eine Funktion f ist fallend in einen bestimmten Intervall, falls für alle \displaystyle x_1 und \displaystyle x_2 im Intervall:
\displaystyle x_1 < x_2\quad\Rightarrow\quad f(x_1) \ge f(x_2)\,\mbox{.} |
Die Definition sagt uns also dass ein Punkt rechts von einen bestimmten Punkt, immer einen höheren oder zumindest denselben Funktionswert hat als der linke Punkt. Laut der Definition kann eine konstante Funktion gleichzeitig steigend und fallend sein.
Da dies manchmal unerwünscht ist, definiert man die Begriffe streng steigend und streng fallend:
Eine Funktion f ist streng steigend in einen bestimmten Intervall, falls für alle \displaystyle x_1 und \displaystyle x_2 im Intervall:
\displaystyle x_1 < x_2\quad\Rightarrow\quad f(x_1) < f(x_2)\,\mbox{.} |
Eine Funktion f ist streng fallend in einen bestimmten Intervall, falls für alle \displaystyle x_1 und \displaystyle x_2 im Intervall:
\displaystyle x_1 < x_2\quad\Rightarrow\quad f(x_1) > f(x_2)\,\mbox{.} |
(Eine streng steigende oder fallend Funktion kann nicht konstant sein)
Beispiel 1
- Die Funktion \displaystyle y= f(x) whose graph is given in the chart below on the far left is increasing in the interval \displaystyle 0 \le x \le 6.
- Die Funktion \displaystyle y=-x^3\!/4 is a strictly decreasing function.
- Die Funktion \displaystyle y=x^2 is a strictly increasing function for \displaystyle x \ge 0.
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Graph of Die Funktion in part a | Graph of Die Funktion f(x) = - x³/4 | Graph of Die Funktion f(x) = x² |
The derivative may of course be used to examine whether a function is increasing or decreasing. We have that
\displaystyle \begin{align*} f^{\,\prime}(x) > 0 \quad&\Rightarrow \quad f(x) \text{ is (strictly) increasing,}\\ f^{\,\prime}(x) < 0 \quad&\Rightarrow \quad f(x) \text{ is (strictly) decreasing.} \end{align*} |
Note, however, that this works only one way round. It is entirely possible for a function to be strictly increasing or decreasing on some interval, and for there to be a point, or perhaps more than one, within that interval at which the derivative is zero. As long as any points where the derivative is zero are isolated (that is, provided the derivative isn't zero anywhere close to such points, but only at them), then Die Funktion can be strictly increasing or decreasing; problems only arise if the gradient is zero, and Die Funktion therefore constant, over some interval.
Stationary points
Points where \displaystyle f^{\,\prime}(x) = 0 are known as stationary points (the term "critical points" is also sometimes used). They are usually one of three kinds:
- Local maximum with \displaystyle f^{\,\prime}(x) > 0 to the left, and \displaystyle f^{\,\prime}(x) < 0 to the right of the point.
- Local minimum with \displaystyle f^{\,\prime}(x) < 0 to the left, and \displaystyle f^{\,\prime}(x) > 0 to the right of the point.
- Stationary point of inflexion with \displaystyle f^{\,\prime}(x) < 0 or \displaystyle f^{\,\prime}(x) > 0 on both sides of the point.
Note that a point may be a local maximum or minimum without \displaystyle f^{\,\prime}(x) = 0; learn more about this in the section on maxima and minima.
Points of inflexion
A point of inflexion is a point where the direction of curvature of a graph changes: where, having curved upwards, it begins curving downwards, or vice versa. In other words, it is a local maximum or minimum for the gradient.
One way to think about points of inflexion is to imagine driving a car along a road shaped like the curve; a point of inflexion is any point at which your steering wheel is exactly centred and you are, for that instant, steering neither right nor left.
At a point of inflexion, the curve's direction of curvature changes; the curve on the left has a stationary point of inflexion in x = 0, where the gradient is zero, but as the other two curves show, not all points of inflexion are stationary points. |
We will not study points of inflexion in depth in this section, except to note that a point of inflexion need not necessarily also be a stationary point (though the two can coincide, in which case we would often call the point concerned a "stationary point of inflexion").
Die Funktion in the above figure has a local minimum at \displaystyle x = -2, a stationary point of inflexion at \displaystyle x = 0 and a local maximum for \displaystyle x = 2.
Table of signs
By studying the derivative sign (+, - or 0), we can therefore obtain a good idea of the curve's appearance.
One creates a so called table of signs. One first determines the x-values where \displaystyle f^{\,\prime}(x) =0 (together with any where the gradient is undefined) and then calculates the sign of the derivative on either side of these points. With the help of some other "backup" points on the curve and using the table of signs one usually can obtain a satisfactory sketch of the curve.
Beispiel 2
Make a table of signs of the derivative of Die Funktion \displaystyle f(x) = x^3 -12x + 6 and then sketch the graph of Die Funktion.
Die Funktions derivative is given by
\displaystyle
f^{\,\prime}(x) = 3x^2 -12 = 3(x^2-4) = 3(x-2)(x+2). |
The factor \displaystyle x-2 is negative to the left of \displaystyle x=2 and positive to the right of \displaystyle x=2. In the same way the factor \displaystyle x+2 is negative to the left of \displaystyle x=-2 and positive to the right of \displaystyle x=-2. This information can be summarised in a table:
\displaystyle x | \displaystyle -2 | \displaystyle 2 | |||
\displaystyle x-2 | \displaystyle - | \displaystyle - | \displaystyle - | \displaystyle 0 | \displaystyle + |
\displaystyle x+2 | \displaystyle - | \displaystyle 0 | \displaystyle + | \displaystyle + | \displaystyle + |
Since the derivative is the product of \displaystyle x-2 and \displaystyle x+2 we thus can determine the derivative sign on the basis of the sign of these factors and create the following table of signs for the derivative on the real-number axis :
\displaystyle x | \displaystyle -2 | \displaystyle 2 | |||
\displaystyle f^{\,\prime}(x) | \displaystyle + | \displaystyle 0 | \displaystyle - | \displaystyle 0 | \displaystyle + |
\displaystyle f(x) | \displaystyle \nearrow | \displaystyle 22 | \displaystyle \searrow | \displaystyle -10 | \displaystyle \nearrow |
In the table's last line, we have given arrows that indicate whether Die Funktion is strictly increasing \displaystyle (\,\nearrow\,\,) or strictly decreasing \displaystyle (\,\searrow\,\,) in each interval as well as the value of Die Funktion value at the stationary points \displaystyle x=-2 and \displaystyle x=2.
From the figure, we see that \displaystyle f(x) has a local maximum at \displaystyle (–2, 22) and a local minimum at \displaystyle (2, –10). The graph now can be sketched:
Maxima and minima (extrema)
A point at which a function takes on its largest or smallest value in comparison with its immediate surroundings is called a local maximum or local minimum (often abbreviated to max and min). Local maxima and minima are together known as extrema.
An extremum may occur in one of three ways:
- At a stationary point (where \displaystyle f^{\,\prime}(x)=0\,).
- At a point where the derivative does not exist (known as a singular point).
- At an endpoint to the interval where Die Funktion is defined.
Beispiel 3
For Die Funktion below there are four extrema: maximum at \displaystyle x=c and \displaystyle x=e, and minimum at \displaystyle x=a and \displaystyle x=d.
At \displaystyle x=a, \displaystyle x=b and \displaystyle x=d one has \displaystyle f^{\,\prime}(x) =0, but it is only at \displaystyle x=a and \displaystyle x=d that there are extrema, since \displaystyle x=b is a stationary point of inflexion.
At \displaystyle x=c the derivative is not defined (as it is a cusp or corner of the curve and it is not possible to determine the slope). The point \displaystyle x=e is an endpoint.
When one is looking for the extrema of a function one must discover and examine all possible candidates for these points. An appropriate working procedures is:
- Differentiate Die Funktion.
- Check to see if there are any points where \displaystyle f^{\,\prime}(x) is not defined.
- Determine all points where \displaystyle f^{\,\prime}(x) = 0.
- Make a table of signs to locate and classify all of the extrema.
- Calculate the value of Die Funktion for all the extrema and at any endpoints.
Beispiel 4
Determine all the extrema of the curve \displaystyle y=3x^4 +4x^3 - 12x^2 + 12.
Die Funktion's derivative is given by
\displaystyle
y' = 12x^3 + 12x^2 - 24x = 12x(x^2+x-2)\,\mbox{.} |
In order to determine how the sign of the derivative varies along the real-number axis, we factorise the derivative as completely as possible. We have already managed to take out the factor \displaystyle 12x and we can factorise further the remaining term \displaystyle x^2+x-2 by identifying its zeros
\displaystyle
x^2+x-2=0\qquad\Leftrightarrow\qquad x=-2\quad\text{or}\quad x=1. |
This means that \displaystyle x^2+x-2=(x+2)(x-1) and the derivative can be rewritten as
\displaystyle y' = 12x(x+2)(x-1)\,\mbox{.} |
It can be seen immediately from this that the derivative is zero for \displaystyle x=-2, \displaystyle x=0 and \displaystyle x=1. In addition, we can see how the derivatives sign varies by examining the sign of each individual factor in the product for different values of \displaystyle x.
\displaystyle x | \displaystyle -2 | \displaystyle 0 | \displaystyle 1 | ||||
\displaystyle x+2 | \displaystyle - | \displaystyle 0 | \displaystyle + | \displaystyle + | \displaystyle + | \displaystyle + | \displaystyle + |
\displaystyle x | \displaystyle - | \displaystyle - | \displaystyle - | \displaystyle 0 | \displaystyle + | \displaystyle + | \displaystyle + |
\displaystyle x-1 | \displaystyle - | \displaystyle - | \displaystyle - | \displaystyle - | \displaystyle - | \displaystyle 0 | \displaystyle + |
The derivative is the product of these factors, and we may obtain the sign of the derivative by multiplying together signs of the factors in each interval.
\displaystyle x | \displaystyle -2 | \displaystyle 0 | \displaystyle 1 | ||||
\displaystyle f^{\,\prime}(x) | \displaystyle - | \displaystyle 0 | \displaystyle + | \displaystyle 0 | \displaystyle - | \displaystyle 0 | \displaystyle + |
\displaystyle f(x) | \displaystyle \searrow | \displaystyle -20 | \displaystyle \nearrow | \displaystyle 12 | \displaystyle \searrow | \displaystyle 7 | \displaystyle \nearrow |
The curve has thus local minima at \displaystyle (–2, –20) and \displaystyle (1, 7) and a local maximum at \displaystyle (0, 12).
Beispiel 5
Determine all extrema of the curve \displaystyle y= x - x^{2/3}.
The derivative of Die Funktion is given by
\displaystyle
y' = 1 - \frac{2}{3} x^{-1/3} = 1- \frac {2}{3} \, \frac{1}{\sqrt[\scriptstyle 3]{x}}\,\mbox{.} |
From this expression, we see that \displaystyle y' is not defined for \displaystyle x = 0 (although which \displaystyle y is defined). This means that Die Funktion has a singular point at \displaystyle x=0.
The stationary points of Die Funktion are given by
\displaystyle
y'=0 \quad \Leftrightarrow \quad 1= \frac {2}{3} \, \frac{1}{\sqrt[3]{x}}\quad\Leftrightarrow\quad \sqrt[3]{x} = \tfrac {2}{3}\quad \Leftrightarrow \quad x = \bigl(\tfrac{2}{3}\bigr)^3 = \tfrac{8}{27}\,\mbox{.} |
The only points at which Die Funktion might have an extremum are thus \displaystyle x=0 and \displaystyle x=\tfrac{8}{27}. In order to determine the nature of these points we create a table of signs:
\displaystyle x | \displaystyle 0 | \displaystyle \frac{8}{27} | |||
\displaystyle y' | \displaystyle + | not def. | \displaystyle - | \displaystyle 0 | \displaystyle + |
\displaystyle y | \displaystyle \nearrow | \displaystyle 0 | \displaystyle \searrow | \displaystyle -\frac{4}{27} | \displaystyle \nearrow |
The curve has a local maximum at \displaystyle (0, 0) (a cusp) and a local minimum at \displaystyle (\tfrac{8}{27},-\tfrac{4}{27})\,.
Global min / max
A function has aglobal maximum at a point if its value there is greater than, or at least equal to, its value at any other point where it is defined; similarly, a global minimum is a point where Die Funktion's value is less than, or at most equal to, its value anywhere else.
To determine a function's global max or min one must therefore find all the extrema and calculate the values of Die Funktion at them. If Die Funktion is defined on an interval with endpoints, one must of course also examine its value at these points.
Note that a function need not have a global max or a global min, even if it has several local extrema.
Beispiel 6
In the first figure Die Funktion has no global maximum nor global minimum. In the second figure the function has no global minimum.
In applications, circumstances often dictate that a function has a limited interval where it is defined, i.e. one only studies part of the graph of Die Funktion. One must therefore be careful in case the global max or min is at an endpoint of the interval.
The above function is only of interest in the interval \displaystyle a\le x \le e. We see that the minimum value of Die Funktion in this interval occurs at the stationary point \displaystyle x=b, while the maximim value is found at the endpoint \displaystyle x=e.
Beispiel 7
Determine the maximum and minimum value of Die Funktion \displaystyle f(x) = x^3 -3x + 2 in the interval \displaystyle -0\textrm{.}5 \le x \le 1\,.
We differentiate Die Funktion, \displaystyle f^{\,\prime}(x) = 3x^2 -3, and put the derivative equal to zero to obtain all the stationary points
\displaystyle f^{\,\prime}(x) = 0 \quad \Leftrightarrow \quad x^2 = 1 \quad \Leftrightarrow \quad x= \pm 1\,\mbox{.} |
The point \displaystyle x = –1 is outside the interval on which Die Funktion is defined, and \displaystyle x = 1 lies at one endpoint of this interval. Since Die Funktion has no singular points (it is differentiable everywhere), its maximum and minimum must be at the interval's endpoints,
\displaystyle \begin{align*} f(-0\textrm{.}5) &= 3\textrm{.}375\,\mbox{,}\\[4pt] f(1)&=0\,\mbox{.} \end{align*} |
Die Funktion's maximum value on the given interval is thus \displaystyle 3\textrm{.}375. The minimum value is \displaystyle 0 (see the figure).
The figure shows Die Funktion with the whole graph as a dashed curve , with the part that is within the given interval appearing as a continuous curve.
The second derivative
The sign of the derivative of a function gives us information about whether Die Funktion is increasing or decreasing. Similarly, the sign of the second derivative can show if the first order derivative is increasing or decreasing. This can , among other things, be used to find out whether a given extremum is a maximum or minimum.
If Die Funktion \displaystyle f(x) has a stationary point at \displaystyle x=a where \displaystyle f^{\,\prime\prime}(a)<0, then
- The derivative \displaystyle f^{\,\prime}(x) is strictly decreasing in some interval surrounding \displaystyle x=a.
- Since \displaystyle f^{\,\prime}(a)=0 then \displaystyle f^{\,\prime}(x)>0 to the left of \displaystyle x=a and \displaystyle f^{\,\prime}(x)<0 to the right of \displaystyle x=a.
- This means that Die Funktion \displaystyle f(x) has a local maximum at \displaystyle x=a.
If the derivative is positive to the left of x = a and negative to the right of x = a the function has a local maximum at x = a. |
If Die Funktion \displaystyle f(x) has a stationary point at \displaystyle x=a where \displaystyle f^{\,\prime\prime}(a)>0, then
- The derivative \displaystyle f^{\,\prime}(x) is strictly increasing in some interval around \displaystyle x=a.
- Since \displaystyle f^{\,\prime}(a)=0 then \displaystyle f^{\,\prime}(x)<0 to the left of \displaystyle x=a and \displaystyle f^{\,\prime}(x)>0 to the right of \displaystyle x=a.
- This means that Die Funktion \displaystyle f(x) has a local minimum at \displaystyle x=a.
If the derivative is negative to the left of x = a and positive to the right of x = a the function has a local minimum at x = a. |
If \displaystyle f^{\,\prime\prime}(a)=0, no information can be deduced, and further investigation is required, for example by means of a table of signs. Note in particular that \displaystyle f^{\,\prime\prime}(a)=0 does not imply that the point is a stationary point of inflexion (although \displaystyle f^{\,\prime\prime}(a)=0 at all points of inflexion, it can also be zero elsewhere, including at maxima and minima).
Beispiel 8
Determine all the extrema of Die Funktion \displaystyle f(x)=x^3 -x^2 -x +2 and determine their character by using the second derivative.
This function is a polynomial and is therefore differentiable everywhere. If Die Funktion has any extrema, they must therefore be found among the stationary points. We thus differentiate Die Funktion, \displaystyle f^{\,\prime}(x) = 3x^2 -2x - 1, and equate the derivative to zero
\displaystyle
f^{\,\prime}(x) = 0 \quad \Leftrightarrow \quad x^2 - \tfrac{2}{3} x - \tfrac{1}{3} = 0 \quad \Leftrightarrow \quad x=1 \quad\text{or}\quad x = -\tfrac{1}{3}\,\mbox{.} |
Die Funktion has stationary points at \displaystyle x = 1 and \displaystyle x=-\tfrac{1}{3}. By examining the sign of the second derivative \displaystyle f^{\,\prime\prime}(x)=6x-2, we can classify each stationary point .
- For \displaystyle x=-\tfrac{1}{3} we have that \displaystyle f^{\,\prime\prime}(-\tfrac{1}{3})=-4<0 and that means that \displaystyle x=-\tfrac{1}{3} is a local maximum.
- For \displaystyle x=1 we have that \displaystyle f^{\,\prime\prime}(1)=4>0 and that means that \displaystyle x=1 is a local maximum.