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Mathematical Expression Editor
We examine a fact about continuous functions.
Let’s recall the idea of a global extrema.
(a)
A function \(f\) has a global maximum at \(x=a\), if \(f(a)\geq f(x)\) for every \(x\) in the domain of
the function.
(b)
A function \(f\) has a global minimum at \(x=a\), if \(f(a)\leq f(x)\) for every \(x\) in the domain of
the function.
A global extremum is either a global maximum or a global minimum.
Let \(f\) be the function given by the graph below.
Find
the \(x\)-coordinate of the point where the function \(f\) has a global maximum.
\[ x= \answer [given]{2} \]
Observe that \(f(2)\ge f(x)\) for all \(x\) in the domain of \(f\). Notice that the function \(f\) has also a local
maximum at \(x=2\).
Find the \(x\)-coordinate of the point where the function \(f\) has a global minimum.
\[ x= \answer [given]{5} \]
Observe that \(f(5)\le f(x)\) for all \(x\) in the domain of \(f\). Notice that the function \(f\) does not have a
local minimum at \(x=5\). Recall, a function cannot not have a local extremum at a
boundary point.
Find the \(x\)-coordinate(s) of the point(s) where the function \(f\)
has a local minimum.
\[ x= \answer [given]{1} \]
Observe that \(f(1)\le f(x)\) for all \(x\) in the interval, say, \((0,2)\). But it is not
true that \(f(1)\le f(x)\) for all \(x\) in the domain of \(f\). For example, \(f(4.5)<f(1)\).
Does every
function attain a global extremum on its domain? Select the correct answer.
All functions must attain both global minimum and global maximum on their
domain. All functions attain a global minimum on their domain. All functions
attain a global maximum on their domain. Some functions have no global
extremums on their domain.
Check the following graph.
Notice, the function is not continuous at \(x=1\), and, therefore, \(f\) is not continuous on
its domain, \((-1,5)\). Does the function given by the graph above attain a global extremum on its domain?
Select the correct answer.
The function attains both global minimum and global maximum on its domain. The function attains a global minimum, but has no global maximum on its domain. The function attains a global maximum, but has no global minimum on its domain. The function has no global extremum on its domain.
Check the graph below.
Notice, the
function is continuous on its domain \((-1,5)\). Does the function given by the graph
above attain a global extremum on its domain? Select the correct answer.
The function attains both global minimum and global maximum on its domain.
The function attains a global minimum, but has no global maximum on its domain. The function attains a global maximum, but has no global minimum on its domain. The function has no global extremum on its domain.
Check the following graph.
Notice, the
function is continuous on a closed interval\([-1,5]\). Does the function given by the graph
above attain a global extremum on its domain? Select the correct answer.
The function attains both global minimum and global maximum on its domain. The function attains a global minimum, but has no global maximum on
its domain. The function attains a global maximum, but has no global
minimum on its domain. The function has no global extremum on its domain.
Find the x-coordinate(s) of the point(s) where the function \(f\) has a global minimum.
Find the x-coordinate(s) of the point(s) where the function \(f\) has a global
maximum.
\[ x= \answer [given]{5} \]
Sometimes it is important to know whether a function attains a
global extremum on its domain. The following theorem, which comes as no surprise
after the previous three examples, gives a simple answer to that question.
Extreme
Value Theorem If \(f\) is continuous on the closed interval \([a,b]\), then there are points
\(c\) and \(d\) in \([a,b]\), such that \((c,f(c))\) is a global maximum and \((d,f(d))\) is a global minimum on
\([a, b]\).
Below, we see a geometric interpretation of this theorem.
Would this theorem hold if we were working on an open interval?
yes no
Consider \(\tan (\theta )\) for \(-\pi /2 < \theta < \pi /2\). Does this function achieve its maximum or minimum?
Would this theorem hold if we were working on a closed interval \([a,b]\), but a function is
not continuous on \([a,b]\)?
yes no
Consider a function \(f\) on a closed interval \([-1,1]\),
defined by \(f(x)=\frac {1}{x}\) for \(x\neq 0\) and \(f(0)=0\) . Does this function achieve its maximum or minimum?
Assume that a function \(f\) is continuous on a closed interval \([a,b]\). By the Extreme Value
Theorem, \(f\) attains both global extremums on the interval \([a,b]\). How can we locate these
global extrema? We have seen that they can occur at the end points or in the
open interval \((a,b)\). If a global extremum occurs at a point \(x\) in the open interval \((a,b)\),
then \(f\) has a local extremum at \(x\). That means that \(f\) has a critical point at \(x\).
So, the global extrema of a function \(f\) occur either at the end points, \(a\) or \(b\),
or at critical points. If we want to locate the global extrema, we have to
evaluate the function at the end points and at critical points, and compare the
values.
Find the extreme values of \(f(x) = x^3 - 3x^2 - 1\) on the interval \([-1,4]\).
The critical points are \(x=0\) and \(x=2\) (both these lie in \((-1,4)\)), with values \(f(0) = -1\) and \(f(2) = -5\).
The endpoints have values \(f(-1) = -5\) and \(f(4) = 15\).
That gives us a list of values \(\{ -1, -5, 15\}\).
The global maximum value of \(f\) is \(15\) (occurring at \(x=4\)) and a global minimum value of \(-5\)
(occurring at \(x=-1\) and \(x=2\)).
Let \(f(x)=x^2e^{-x}\), for \( -2\le x\le 1\). Locate the global extremums of \(f\) on the closed interval \([-2,1]\). Does the function \(f\)
satisfy the conditions of the Extreme Value Theorem on its domain?
Yes, because \(f\)
is continuous on \((-2,1)\).Yes, because \(f\) is continuous on \([-2,1]\).No, because \(f\) is not continuous
on \([-2,1]\).No, because \(f\) is not differentiable on \((-2,1)\).
Therefore, the Extreme Value
Theorem guarantees that the function \(f\) attains both global extremums on
its domain. The global extremums occur at the end points or at critical
points.
Find the critical points of \(f\). First, compute the derivative of \(f\).
\[ f'(x)= xe^{-x}(\answer [given]{2-x}) \]
In order to find
the critical points of \(f\), we have to solve the equation
\[ f'(x)= \answer [given]{0}. \]
It follows that the
function \(f\) has only one critical point \(\left (c,f(c)\right )\). Find \(c\).
\[ c= \answer [given]{0} \]
In order to locate the global
extremums of \(f\), we have to evaluate \(f\) at the end points and at the critical point.
\[ f(-2)= \answer [given]{4e^{2}} \]
\[ f(c)= \answer [given]{0} \]
\[ f(1)= \answer [given]{e^{-1}} \]
Order the three values, \(f(-2)\), \(f(c)\), and \(f(1)\), from smallest to largest. You should replace \(c\) with
its value, when you write \(f(c)\) in your answer below.
Based on this comparison, find the location of the global minimum and global
maximum of \(f\). Circle the correct answer.
\(f\) has a global maximum at \(x=1\) and
global minimum at \(x=0\).\(f\) has a global maximum at \(x=1\) and global minimum at \(x=-2\).\(f\) has a global maximum at \(x=-2\) and global minimum at \(x=0\).\(f\) has a global
maximum at \(x=-2\) and global minimum at \(x=1\).\(f\) has a global maximum at \(x=0\) and
global minimum at \(x=-2\).\(f\) has a global maximum at \(x=0\) and global minimum at \(x=1\).
Find the extreme values of \(g(x) = \sin (x) - \cos (x)\) on \([-\pi , \pi ]\).
Notice that \(x\)-values with \(\cos (x) = 0\) have \(\sin (x) = \pm 1\), so \(g'(x) = 0 \pm 1 \neq 0\). These values are not critical points.
In the interval \([-\pi , \pi ]\), the solutions of \(\tan (x) = -1\) are \(x = -\frac {\pi }{4}, \frac {3\pi }{4}\). (The solutions not in \((-\pi , \pi )\) are ignored.)
The endpoints have values \(g( -\pi ) = g(\pi ) = \answer [given]{1}\)
The global maximum value of \(g\) is \(\answer [given]{\sqrt {2}}\) (occurring at \(x=\answer [given]{\frac {3\pi }{4}}\)) and a global minimum value of \(\answer [given]{-\sqrt {2}}\)
(occurring at \(x=\answer [given]{\frac {\pi }{4} }\) ).