The highest value of a part of a graph is called a local maximum.
The lowest value of a part of a graph 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.
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 maximums and minimums 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 point on a part of the graph. The global maximums and minimums are the highest 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 maximums and minimums, there may not always be a global maximum or minimum.
Using the derivative, we can easily calculate the extreme values of a function.
If a function #\blue{f(x)}# has a local maximum or minimum at #x=\orange{c}# then #\green{f'(\orange{c})}=0#.
Example
\[\begin{array}{rcl}\blue{f(x)}&=&\blue{x^2}\\ \green{f'(x)}&=&\green{2x}\\ \green{f'(\orange{0})}&=&0\end{array}\]
The function must also be differentiable in point #x=\orange{c}#. In this course we will not look at functions that are not differentiable, but this can occur in practice.
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.
This theorem does not hold if #c# is a boundary value of the domain the function is defined on. For example, if we take #f(x)=x^2# on the domain #\ivcc{2}{5}#, then #2# and #5# are extreme values, but the derivative does not equal #0# in those points.
The statment only holds one way. If the derivative #\green{f'(x)}# of a function #\blue{f(x)}# equals #0# in a point #\orange{c}#, this does not immediately mean that #\blue{f(x)}# has an extreme value in #\orange{c}#. For example, the derivative of #\blue{f(x)=x^3}# is equal to #0# in the point #\orange{0}#, but does obviously not have an extreme value here.

Stepbystep

Example


Determine the extreme values of a function #f(x)#. Determine for each extreme value whether it is a local minimum, local maximum or neither.

#\qqquad \begin{array}{rcl}f(x)\phantom{'}&=&x^42x^2\end{array}#

Step 1 
Calculate the derivative #f'(x)#.

#\qqquad \begin{array}{rcl}f'(x)&=&4x^34x\end{array}#

Step 2 
Solve #f'(x)=0# to find the #x# coordinates of the points which are possibly an extreme value.

#\qqquad \begin{array}{rcl} 4x^34x&=&0\\ x&=&0 \lor 4x^24=0\\ x&=&0\lor x^2=1\\ \green{x}&\green{=}&\green{0} \lor \blue{x=1} \lor\orange{x=1}\end{array}#

Step 3 
Sketch the graph to find out which points are a local maximum and which points are a local minimum (and which points may not be a maximum or minimum).


Step 4 
Substitute the obtained #x# coordinates in #f(x)# and determine the extreme values this way.

#f(\blue{1})=1#, #f(\orange{1})=1#, #f(\green{0})=0#
Therefore, local minimum is #1# and local maximum is #0#

Calculating extrema of functions is something which appears often in optimization 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# metres 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# metres of fence, which has to be distributed over the width and depth, which gives us #2x+2y=500#. Rearranging gives \[y=250x\] We insert this in the function for the are and get \[A=x\cdot(250x) = 250\cdot x  x^2\] This is the formule whose maximum we would like to calculate. Following the stepbystep 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 contains 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 a interval are all positive or all 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_{i1}}{x_i}# and #\ivoo{x_i}{x_{i+1}}#.
 If the sign of #\ivoo{x_{i1}}{x_i}# is positive and the sign of #\ivoo{x_i}{x_{i+1}}# is negative, then #x_i# corresponds to a local maximum.
 If the sign of #\ivoo{x_{i1}}{x_i}# is negative and the sign of #\ivoo{x_i}{x_{i+1}}# is positive, then #x_i# corresponds to a local minimum.
 If the sign of #\ivoo{x_{i1}}{x_i}# is positive and the sign of #\ivoo{x_i}{x_{i+1}}# is positive, then #x_i# does not correspond to an extreme value.
 If the sign of #\ivoo{x_{i1}}{x_i}# is negative and the sign of #\ivoo{x_i}{x_{i+1}}# is negative, then #x_i# does not correspond to an extreme value.
In the example we found the zeroes #x_1=1, x_2=0# and #x_3=1#. 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 with 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 \[f(x)=x^34x^2+4x+1\] 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_=# #{{2}\over{3}}# and #x_+=# #2#
We start by noting that the domain of #f# is all real numbers, so extreme values can only occur in stationary points.
Step 1 
We determine the derivative of #f(x)=x^34x^2+4x+1#. This is equal to: \[f'(x)=3x^28x+4\] 
Step 2 
We determine the #x#coordinates of the stationary points by setting the derivative equal to #0# and solving the following equation \[\begin{array}{rcl}3x^28x+4&=&0 \\ &&\phantom{xxx}\blue{\text{the equation we need to solve}}\\ x=\frac{8\sqrt{(8)^24\cdot 3\cdot 4}}{2\cdot 3} &\lor& x=\frac{8+\sqrt{(8)^24\cdot 3\cdot 4}}{2\cdot 3} \\ &&\phantom{xxx}\blue{\text{quadratic formula}}\\ x=\frac{8\sqrt{16}}{6} &\lor& x=\frac{8+\sqrt{16}}{6} \\ &&\phantom{xxx}\blue{\text{simplified}}\\ x=\frac{84}{6} &\lor& x=\frac{8+4}{6} \\ &&\phantom{xxx}\blue{\text{simplified}}\\ x={{2}\over{3}} &\lor& x=2 \\ &&\phantom{xxx}\blue{\text{simplified}}\end{array}\] Since the derivative of #f# is equal to zero in #x={{2}\over{3}}# and #x=2#, we have that these points are stationary points, and thus the only candidates for extremal values. 
Step 3 
We draw the graph of #f(x)#.
Therefore, there is a local maximum at #x={{2}\over{3}}# and a local minimum at #x=2#. Hence, both obtained #x#values are part of an extreme value. Therefore, #x_={{2}\over{3}}# and #x_+=2# 