The highest value of a part of a graph, which often is a point where the function stops increasing and starts decreasing, is called a local maximum.
The lowest value of a part of a graph, which often is a point where the function stops decreasing and starts increasing, is called a local minimum.
Both are extreme values of a function.
If we view a function on a restricted domain, the values on the boundaries of the domain can also be a local maxima or minima and thus an extreme value. This is the reason we use the word "often" in the above definition.
For example, we can view the function #f(x)=x^2# on the domain #\ivcc{2}{5}#. We now have a local minimum at #x=2# and a local maximum at #x=5#. When #x^2# is considered on its normal domain, we only have a local minimum, at #x=0#.
So far in this course, we often specifically looked at the #x# values of special points, but the maxima and minima are the #y# values of these points.
Thus, in the example, a local maximum of the green graph is #3.5# and #\red{\text{not}}# #0#. A local minimum of the blue graph is #0.5# and #\red{\text{not}}# #0#.
We have seen that the local maxima and minima are the highest/lowest point on a part of the graph. The global maxima and minima are the highest/lowest point of the whole graph.
In the example of the green graph, the local maximum is also global. Similarly, in the blue graph, the local minimum is also global. This is not always the case.
Even if there are local maxima and minima, there may not always be a global maximum or minimum.
For example, look at the graph of function #f(x)=x^4-x^3-5x^2+5#. This graph has local minima at #x=-1.25# and #x=2#, of which the one at #x=2# is also a global minimum. The graph also has a local maximum at #x=0#, and no global maximum.
We want to be able to find the extreme values of a function. To do this, we first define the stationary points of a function.
Point #x=\orange{c}# is a stationary point if #\green{f'(}\orange{c}\green{)}=0#.
Example
\begin{array}{rcl}\blue{f(x)}&=&\blue{x^2-2x}\\ \green{f'(x)}&=&\green{2x-2}\\ \green{f'(}\orange{1}\green{)}&=&0\end{array} so #x=\orange{1}# is a
stationary point.
If the derivative in a point equals zero, then the tangent line to the function is horizontal, as we can see in the following picture.
Using this definition, we can find the extreme values of a function.
If #\blue{f(x)}# has an extreme value at #x=\orange{c}#, then
#\bullet \enspace x=\orange{c}# is a stationary point, or
#\bullet \enspace x=\orange{c}# is the boundary of the domain of #\blue{f(x)}#.
Example
#\blue{f(x)}=\blue{\sqrt{x}}# has a local minimum at #x=\orange{0}#.
This is the boundary of the domain of #\blue{f(x)}#, which is #\ivco{0}{\infty}#.
The statement above is a slightly simplified version of the following statement:
If #\blue{f(x)}# has an extreme value at #x=\orange{c}#, then #x=\orange{c}# is a
critical point, where critical points are points such that:
#\bullet \enspace x=\orange{c}# is a stationary point, so #\green{f'(}\orange{c}\green{)}=0#, or
#\bullet \enspace x=\orange{c}# is the boundary of the domain of #\blue{f(x)}#, or
#\bullet \enspace \green{f'(}\orange{c}\green{)}# doesn't exist.
Note that we left out the last bullet point in the simplified statement.
As an example, consider #\blue{f(x)}=\blue{|x|}# at #x=\orange{0}#. We have that #\green{f'(}\orange{0}\green{)}# does not exist, but still this function has a local (and global) minimum in #x=\orange{0}#.
Functions for which an extreme value occurs at a point in which the derivative is nonexistent are outside the scope of this course.
Below, we explain how to find the extreme values of a function step by step.
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Step-by-step
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Example
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Find all extremal values of a function #f(x)#.
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#\qqquad \begin{array}{rcl}f(x)\phantom{'}&=&x^4-2x^2\end{array}#
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| Step 1 |
Calculate the derivative #f'(x)#.
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#\qqquad \begin{array}{rcl}f'(x)&=&4x^3-4x\end{array}#
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| Step 2 |
Solve #f'(x)=0# to find the stationary points of #f(x)#.
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#\qqquad \begin{array}{rcl} 4x^3-4x&=&0\\ x&=&0 \lor 4x^2-4=0\\ x&=&0\lor x^2=1\\ \green{x}&\green{=}&\green{0} \lor \blue{x=-1} \lor\orange{x=1}\end{array}#
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| Step 3 |
Write down all stationary points and points on the boundary of the domain of #f(x)# (if there are any).
Then sketch the graph of #f(x)# to find out which points are a local maximum and which points are a local minimum (and which points are neither).
If it is not clear how to sketch the graph, it is better to make a sign analysis chart.
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| Step 4 |
Substitute the obtained #x# coordinates in #f(x)# and determine the extreme values this way.
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#f(\blue{-1})=-1\,#,
#f(\orange{1})=-1\,#,
#f(\green{0})=0#
Therefore, there are two local minima at #x=\pm 1#, both with value #-1#, and a
local maximum at #x=0# with value #0#.
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Calculating the extrema of functions is something that appears often in optimisation problems. Problems are described by functions, whose minimum or maximum are determined.
We will give a very easy example. Suppose a farmer wants to fence a rectangular field, and he has bought #500# meters of fence. The farmer wants to maximize the fenced area and wants to know the best ratio of the rectangle. We first note that this area #A# is given by #x\cdot y#, where #x# is the width and #y# is the depth of the rectangle. The farmer bought #500# meters of fence, which has to be distributed over the width and depth, which gives us #2x+2y=500#. Rearranging gives \[y=250-x\] We insert this in the function for the area and get \[A=x\cdot(250-x) = 250\cdot x - x^2\] This is the formula whose maximum we would like to calculate. Following the step-by-step approach, we get #x=125#. Thus, the farmer should fence a square area in order to maximize the area fenced.
In most applications, the functions are very complicated and contain much more variables. However, we won't be studying this in this course.
As an alternative to step #3#, we can make use of a so-called sign analysis chart. In step #2# we found the zeroes #x_1,\ldots, x_n# of the derivative #f'(x)#. By definition, there are no zeroes in the intervals #\ivoo{x_i}{x_{i+1}}#. This means that the values of #f'(x)# in such an interval are all positive or negative. If we take one point in the interval and substitute it in #f'(x)# we immediately know the sign of the interval: positive or negative. We now write the signs down for all the intervals in a sign analysis chart
| Interval |
#\ivoo{-\infty}{x_1}# |
#x_1# |
#\ivoo{x_1}{x_2}# |
#x_2# |
#\ldots# |
#x_n# |
#\ivoo{x_n}{\infty}# |
| Sign |
#+# or #-# |
#0# |
#+# or #-# |
#0# |
#\ldots# |
#0# |
#+# or #-# |
We can use this sign analysis chart to determine whether a zero #x_1# of #f'(x)# corresponds to a local maximum, minimum, or none. This is done by considering the signs of the intervals surrounding it, which are #\ivoo{x_{i-1}}{x_i}# and #\ivoo{x_i}{x_{i+1}}#.
- #x_i# is a minimum if the sign of the derivative goes from negative in the previous interval to positive in the next.
- #x_i# is a maximum if the sign of the derivative goes from positive in the previous interval to negative in the next.
- #x_i# is not an extreme value if the sign of the derivative does not change from the previous interval to the next.
In the example we found the zeroes #x_1=-1, x_2=0# and #x_3=1#. To find the signs in the second, fourth, sixth, and eighth column, we choose the following values:
- For the interval #\ivoo{-\infty}{-1}# we choose for example #x=-2#. We have #f'(-2)=4\cdot (-2)^3-4\cdot (-2)=-24#, which is negative, so we put a minus in the table.
- For the interval #\ivoo{-1}{0}# we choose for example #x=-\frac{1}{2}#. We have #f'\left(-\frac{1}{2}\right)= \frac{3}{2}#, which is positive, so we put a plus in the table.
- For the interval #\ivoo{0}{1}# we choose for example #x=\frac{1}{2}#. We have #f'\left(\frac{1}{2}\right)= -\frac{3}{2}#, which is negative, so we put a minus in the table.
- For the interval #\ivoo{1}{\infty}# we choose for example #x=2#. We have #f'(2)=24#, which is positive, so we put a plus in the table.
This yields the following sign analysis chart.
| Interval |
#\ivoo{-\infty}{-1}# |
#-1# |
#\ivoo{-1}{0}# |
#0# |
#\ivoo{0}{1}# |
#1# |
#\ivoo{1}{\infty}# |
| Sign |
#-# |
#0# |
#+# |
#0# |
#-# |
#0# |
#+# |
We see that both #x=-1# and #x=1# correspond to a local minimum and that #x=0# corresponds with a local maximum.
Give the two values of #x# for which the function #f# given by \[f(x)=x^3-2x^2+x+2\] has an extreme value (a local minimum or local maximum).
The smaller value of #x# is indicated by #x_-# and the greater value by #x_+#. Write your answers as a simplified fraction.
#x_-=# #{{1}\over{3}}# and #x_+=# #1#
| Step 1 |
We determine the derivative of #f(x)=x^3-2x^2+x+2#. This is equal to:
\[f'(x)=3x^2-4x+1\] |
| Step 2 |
We determine the #x# coordinates of the potential extreme values by making the derivative equal to #0# by solving the equation.
\[\begin{array}{rcl}3x^2-4x+1&=&0 \\ &&\phantom{xxx}\blue{\text{the equation we need to solve}}\\
x=\frac{4-\sqrt{(-4)^2-4\cdot 3\cdot 1}}{2\cdot 3} &\lor& x=\frac{4+\sqrt{(-4)^2-4\cdot 3\cdot 1}}{2\cdot 3} \\ &&\phantom{xxx}\blue{\text{quadratic formula}}\\
x=\frac{4-\sqrt{4}}{6} &\lor& x=\frac{4+\sqrt{4}}{6} \\ &&\phantom{xxx}\blue{\text{simplified}}\\
x=\frac{4-2}{6} &\lor& x=\frac{4+2}{6} \\ &&\phantom{xxx}\blue{\text{simplified}}\\
x={{1}\over{3}} &\lor& x=1 \\ &&\phantom{xxx}\blue{\text{simplified}}\end{array}\] |
| Step 3 |
We draw the graph of #f(x)#.
Therefore, there is a local maximum at #x={{1}\over{3}}# and a local minimum at #x=1#. Hence, both obtained #x# values are part of an extreme value.
Therefore, #x_-={{1}\over{3}}# and #x_+=1# |
Extreme values
