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Mathematical Expression Editor
We use derivatives to help locate extrema.
Whether we are interested in a function as a purely mathematical object or in
connection with some application to the real world, it is often useful to know
what the graph of the function looks like. We can obtain a good picture of
the graph using certain crucial information provided by derivatives of the
function.
1 Extrema
We’ll start with some definitions.
(a)
A function \(f\) has an 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 an 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. These are
occasionally referred to as absolute extrema as well.
Graphically, the global extrema correspond to the highest and lowest points on a
graph of a function.
Find the global extrema of the function \(f(x) = x^2\).
Examine the graph of the \(y=x^2\).
There is no highest point on this graph since \(\displaystyle \lim _{x\to \pm \infty }x^2 = \infty \), but there is a lowest point.
The function has a global minimum at \(x=0\) (the vertex of the parabola) and no global
maximum.
Find the global extrema of the function \(f(x) = \cos (x)\).
Examine the graph of \(y=\cos (x)\).
From the graph, \(\cos (x)\) has a highest value of \(1\) that occurs at all even multiples of \(\pi \): at \(x=2\pi k\) for \(k\)
any integer. All of these points are global maxima.
The lowest value of \(-1\) occurs at odd multiples of \(\pi \): at \(x=(2k+1)\pi \) for \(k\) any integer. All of these
points are global minima.
Derivatives will be more helpful when considering not global extrema, but their local
counterparts. A local extremum of a function \(f\) is a point \((a,f(a))\) on the graph of \(f\) where the
\(y\)-coordinate is larger (or smaller) than all other \(y\)-coordinates of points on the graph
whose \(x\)-coordinates are “close to” \(a\).
(a)
A function \(f\) has a local maximum at \(x=a\), if \(f(a)\geq f(x)\) for every \(x\) in some open interval
I containing \(a\).
(b)
A function \(f\) has a local minimum at \(x=a\), if \(f(a)\leq f(x)\) for every \(x\) in some open interval
I containing \(a\).
A local extremum is either a local maximum or a local minimum. These are
sometimes referred to as relative extrema as well.
In our next example, we clarify the definition of a local minimum.
Consider the graph
of a function \(f\):
Identify the local extrema of \(f\) and give an explanation.
In the figure above the
function \(f\) has a local minimum at,
\[ a=\answer [given]{1}, \]
because we can find an open interval \(I\) (marked in
the figure) that contains the number
\[ a=\answer [given]{1}, \]
and for all numbers \(x\) in \(I\) the following statement
is true:
Local maximum and minimum points are quite distinctive on the graph of a
function, and are, therefore, useful in understanding the shape of the graph.
Many problems in real world and in different scientific fields turn out to be
about finding the smallest (or largest) value that a function achieves (for
example, we might want to find the minimum cost at which some task can be
performed).
2 Critical points
Consider the graph of the function \(f\).
The function \(f\) has four local extremums: at \(x=-4\), \(x=-2\), \(x=0\) and \(x=4\). Notice that the function \(f\) is not
differentiable at \(x=-4\) and \(x=-2\). Notice that \(f'(0)=0\) and \(f'(4)=0\).
After this example, the following theorem should not come as a surprise.
Fermat’s Theorem If \(f\) has a local extremum at \(x=a\) and \(f\) is differentiable at \(a\), then \(f'(a)=0\).
Does Fermat’s Theorem say that if \(f'(a) = 0\), then \(f\) has a local extrema at \(x=a\)?
yes no
Consider \(f(x) = x^3\), \(f'(0) = 0\), but \(f\) does not have a local maximum or minimum at \(x=0\).
Fermat’s Theorem says that the only points at which a function can have a local
maximum or minimum are points at which the derivative is zero or the derivative is
undefined. As an illustration of the first scenario, consider the plots of \(f(x) = x^3-4.5x^2+6x\) and \(f'(x) = 3x^2-9x+6\).
Make a
correct choice that completes the sentence below.
At the point \((1,f(1))\), the function \(f\) has
a local maximum a local minimum no local extremum
Select the correct
statement.
\(f'(1)\) is undefined\(f'(1)>0\)\(f'(1)=0\)\(f'(1)<0\)
Make a correct choice that completes the
sentence below.
At the point \((1.5,f(1.5))\), the function \(f\) has
a local maximum a local minimum no local extremum
Select the correct
statement.
\(f'(1.5)\) is undefined\(f'(1.5)>0\)\(f'(1.5)=0\)\(f'(1.5)<0\)
Make a correct choice that completes the sentence below.
At the point \((2,f(2))\), the function \(f\) has
a local maximum a local minimum no local extremum
Select the correct
statement.
\(f'(2)\) is undefined\(f'(2)>0\)\(f'(2)=0\)\(f'(2)<0\)
As an illustration of the second scenario, consider
the plots of \(f(x) = x^{2/3}\) and \(f'(x) = \frac {2}{3x^{1/3}}\):
Make a correct choice that completes the sentence below.
At the point \((-2,f(-2))\), the function \(f\) has
a local maximum a local minimum no local extremum
Select the correct
statement.
\(f'(-2)\) is undefined\(f'(-2)>0\)\(f'(-2)=0\)\(f'(-2)<0\)
Make a correct choice that completes the
sentence below.
At the point \((0,0)\), the function \(f\) has
a local maximum a local minimum no local extremum
Select the correct
statement.
\(f'(0)\) is undefined\(f'(0)>0\)\(f'(0)=0\)\(f'(0)<0\)
This brings us to our next definition.
Assume that a function \(f\) is defined on an open interval I that contains a point \(a\). Then
we say that the function \(f\) has a critical point at \(x=a\) if
\[ f'(a) = 0\qquad \text {or}\qquad \text {$f'(a)$ does not exist.} \]
Notice, if a function \(f\) has a
critical point at \(x=a\), then the number \(a\) is inside some open interval \(I\), and \(I\) is in the
domain of \(f\).
When looking for local maximum and minimum points, be careful not to make two
sorts of mistakes:
You may forget that a maximum or minimum can occur where the
derivative does not exist, and so forget to check whether the derivative
exists everywhere.
You might assume that any place that the derivative is zero is a local maximum
or minimum point, but this is not true, consider the plots of \(f(x) = x^3\) and \(f'(x) = 3x^2\).
While \(f'(0)=0\),
there is neither a maximum nor minimum at \(x=0\).
Since both local maximum and local minimum occur at a critical point, when
we locate a critical point, we need to determine which, if either, actually
occurs.
Find all local maximum and minimum points for the function \(f(x)=x^3-x\).
Write
\[ \ddx f(x)=\answer [given]{3x^2-1}. \]
We can easily express \(f'(x)\) as a product of its factors
which implies that the
function \(f\) has only two critical points, \(x=-\frac {\sqrt {3}}{3}\) and \(x=\frac {\sqrt {3}}{3}\). Notice that the derivative \(f'(x)\)
is a polynomial, and polynomials do not change signs except possibly at
their zeros. This implies that derivative \(f'(x)\)does not change the sign on
the intervals \(\left (-\infty ,-\frac {\sqrt {3}}{3}\right )\), \(\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )\), and \(\left (\frac {\sqrt {3}}{3},\infty \right )\), because these intervals do not contain any zeros of
\(f'(x)\).
Select the correct statement about the sign of \(f'(x)\) on the intervals \(\left (-\infty ,-\frac {\sqrt {3}}{3}\right )\) and \(\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )\).
\(f'(x)>0\)
on \(\left (-\infty ,-\frac {\sqrt {3}}{3}\right )\) and \(f'(x)>0\) on \(\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )\). \(f'(x)>0\) on \(\left (-\infty ,-\frac {\sqrt {3}}{3}\right )\) and \(f'(x)<0\) on \(\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )\). \(f'(x)<0\) on \(\left (-\infty ,-\frac {\sqrt {3}}{3}\right )\) and \(f'(x)>0\) on \(\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )\). \(f'(x)<0\) on \(\left (-\infty ,-\frac {\sqrt {3}}{3}\right )\) and \(f'(x)<0\) on \(\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )\)
If
we know the sign of the derivative on an interval, we also know whether
the function is increasing or decreasing on that interval. This will help us
determine whether the function has a local extremum at the critical point where \(x=-\frac {\sqrt {3}}{3}\).
At the critical point where \(x=-\frac {\sqrt {3}}{3}\), the function \(f\) has
no local extremum, because \(f\) is increasing on \(\left (-\infty ,-\frac {\sqrt {3}}{3}\right )\) and increasing on \(\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )\). a local
maximum, because \(f\) is increasing on \(\left (-\infty ,-\frac {\sqrt {3}}{3}\right )\) and decreasing on \(\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )\). a local minimum,
because \(f\) is decreasing on \(\left (-\infty ,-\frac {\sqrt {3}}{3}\right )\) and increasing on \(\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )\). no local extremum, because \(f\) is
decreasing on \(\left (-\infty ,-\frac {\sqrt {3}}{3}\right )\) and decreasing on \(\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )\).
Select the correct statement about the
sign of \(f'(x)\) on the intervals \(\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )\) and \(\left (\frac {\sqrt {3}}{3},\infty \right )\).
\(f'(x)>0\) on \(\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )\) and \(f'(x)>0\) on \(\left (\frac {\sqrt {3}}{3},\infty \right )\).\(f'(x)>0\) on \(\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )\) and \(f'(x)<0\) on \(\left (\frac {\sqrt {3}}{3},\infty \right )\).\(f'(x)<0\) on \(\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )\) and \(f'(x)>0\) on \(\left (\frac {\sqrt {3}}{3},\infty \right )\).\(f'(x)<0\) on \(\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )\) and \(f'(x)<0\) on \(\left (\frac {\sqrt {3}}{3},\infty \right )\).
Again, the sign
of the derivative on an interval determines whether the function is increasing or
decreasing on that interval. This will help us determine whether the function has a
local extremum at the critical point where \(x=\frac {\sqrt {3}}{3}\).
At the critical point where \(x=\frac {\sqrt {3}}{3}\), the function \(f\) has
no local extremum, because \(f\) is increasing on \(\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )\) and increasing on \(\left (\frac {\sqrt {3}}{3},\infty \right )\). a local maximum,
because \(f\) is increasing on \(\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )\) and decreasing on \(\left (\frac {\sqrt {3}}{3},\infty \right )\). a local minimum, because \(f\) is
decreasing on \(\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )\) and increasing on \(\left (\frac {\sqrt {3}}{3},\infty \right )\). no local extremum, because \(f\) is decreasing on\(\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )\)
and decreasing on \(\left (\frac {\sqrt {3}}{3},\infty \right )\).
Do your answers agree with the graphs of \(f\) and \(f'\) given in the
picture below?
3 The first derivative test
We will further explore and refine the method of the previous section for deciding
whether there is a local maximum or minimum at a critical point. Recall that
If \(f'(x) >0\) on an interval, then \(f\) is increasing on that interval.
If \(f'(x) <0\) on an interval, then \(f\) is decreasing on that interval.
So how exactly does the derivative tell us whether there is a maximum, minimum, or
neither at a point? Use the first derivative test.
First Derivative Test Suppose that \(f\) is continuous on an open interval, and that \(f\) has a
critical point at \(x=a\), for some value \(a\) in that interval.
If \(f'(x)>0\) to the left of \(a\) and \(f'(x)<0\) to the right of \(a\), then the function \(f\) has a local
maximum at \(a\).
If \(f'(x)<0\) to the left of \(a\) and \(f'(x)>0\) to the right of \(a\), then the function \(f\) has a local
minimum at \(a\).
If \(f'(x)\) has the same sign to the left and right of \(a\), then the function \(f\) has no
local extremum at \(a\).
Consider the function
\[ f(x) = \frac {x^4}{4}+\frac {x^3}{3}-x^2 \]
Find the intervals on which \(f\) is increasing, the intervals on
which \(f\) is decreasing and identify the local extrema of \(f\).
Start by computing
\[ \ddx f(x) = \answer [given]{x^3+x^2-2x}. \]
Now we need to find on what intervals is \(f'\) positive and what
intervals it is negative. To do this, solve
So the critical points (when \(f'(x)=0\)) are when \(x=-2\), \(x=0\), and \(x=1\). Since the derivative, \(f'(x)\), is a
polynomial, it does not change the sign on intervals between its zeros, i.e.,
between the critical points. Now we can check the sign of \(f'(x)\) at some points
between the critical points to find where \(f'(x)\) is positive and where negative:
Hence \(f\) is increasing on \((-2,0)\) and \((1,\infty )\) and \(f\) is decreasing on \((-\infty ,-2)\) and \((0,1)\). Moreover, from the first
derivative test, the local maximum is at \(x=0\) while the local minimums are at \(x=-2\) and
\(x=1\).
This can be confirmed by checking the graphs of \(f(x) =x^4/4 + x^3/3 -x^2\) and \(f'(x) = x^3 + x^2 -2x\).
Hence we have seen that if \(f'\) is zero at a point and increasing on an interval containing
that point, then \(f\) has a local minimum at the point. If \(f'\) is zero at a point and
decreasing on an interval containing that point, then \(f\) has a local maximum at the
point. Thus, we see that we can gain information about \(f\) by studying how \(f'\) changes.
This leads us to our next section.
4 Inflection points
If we are trying to understand the shape of the graph of a function, knowing where it
is concave up and concave down helps us get a more accurate picture. It is worth
summarizing what we have already seen into a single theorem.
Test for Concavity Suppose that \(f''(x)\) exists on an interval.
(a)
If \(f''(x)>0\) on an interval, then \(f\) is concave up on that interval.
(b)
If \(f''(x)<0\) on an interval, then \(f\) is concave down on that interval.
Of particular interest are points at which the concavity changes from up to down or
down to up.
If \(f\) is continuous at \(x=a\) and its concavity changes either from up to down or down to up
at \(x=a\), then \(f\) has an inflection point at \(x=a\).
It is instructive to see some examples of inflection points:
It is also instructive to see some nonexamples of inflection points:
We identify inflection points by first finding \(x\) such that \(f''(x)\) is zero or undefined and then
checking to see whether \(f''(x)\) does in fact go from positive to negative or negative to
positive at these points.
Even if \(f''(a) = 0\), the point determined by \(x=a\) might not be an inflection point.
Describe the
concavity of \(f(x)=x^4\).
To start, compute the first and second derivative of \(f(x)\) with respect to \(x\),
Since \(f''(0)=0\), there is potentially an inflection point at \(x=0\). But, for any \(x\ne 0\), \(f''(x)>0\). Hence there is no
inflection point at \(x=0\). The curve is concave up for all \(x<0\) and remains concave up for all \(x>0\).
Describe the concavity of \(f(x)=x^3-x\).
To start, compute the first and second derivative of \(f(x)\) with respect to \(x\),
Since \(f''(0)=0\), there is
potentially an inflection point at \(x=0\). Using test points, we note the concavity does
change from down to up, hence there is an inflection point at \(x=0\). The curve
is concave down for all \(x<0\) and concave up for all \(x>0\), see the graphs of \(f(x) = x^3-x\) and \(f''(x) = 6x\).
Note that we need to compute and analyze the second derivative to understand
concavity, so we may as well try to use the second derivative test for maxima
and minima. If for some reason this fails we can then try one of the other
tests.
5 The second derivative test
Recall the first derivative test:
If \(f'(x)>0\) for all \(x\) that are near and to the left of \(a\) and \(f'(x)<0\) for all \(x\) that are near and
to the right of \(a\), then the function \(f\) has a local maximum at \(a\).
If \(f'(x)<0\) for all \(x\) that are near and to the left of \(a\) and \(f'(x)>0\) for all \(x\) that are near and
to the right of \(a\), then the function \(f\) has a local minimum at \(a\).
Assume that a function \(f\) has a critical point at \(a\) and that \(f'(a)=0\). If \(f'\) happens to be
decreasing on some interval containing \(a\), then it changes from positive to negative at \(a\).
Therefore, if \(f''\) is negative on some interval that contains \(a\), then \(f'\) is definitely
decreasing, so there is a local maximum at the point in question. On the other hand,
if \(f'\) is increasing, then it changes from negative to positive at \(a\). Therefore, if \(f''\) is positive
on an interval that contains \(a\), then \(f'\) is definitely increasing, so there is a local
minimum at the point in question. We summarize this as the second derivative
test.
Second Derivative Test Suppose that \(f''(x)\) is continuous on an open interval and that \(f'(a)=0\) for
some value of \(a\) in that interval.
If \(f''(a) <0\), then \(f\) has a local maximum at \(a\).
If \(f''(a) >0\), then \(f\) has a local minimum at \(a\).
If \(f''(a) =0\), then the test is inconclusive. In this case, \(f\) may or may not have a local
extremum at \(x=a\).
The second derivative test is often the easiest way to identify local maximum and
minimum points. Sometimes the test fails and sometimes the second derivative is
quite difficult to evaluate. In such cases we must fall back on one of the previous
tests.
Once again, consider the function
\[ f(x) = \frac {x^4}{4}+\frac {x^3}{3}-x^2 \]
Use the second derivative test to locate the local
extrema of \(f\).
Hence we see that \(f\) has a local
minimum at \(x=-2\), a local maximum at \(x=0\), and a local minimum at \(x=1\), see below for a plot of \(f(x) =x^4/4 + x^3/3 -x^2\)
and \(f''(x) = 3x^2 + 2x -2\):
If \(f''(a)=0\), what does the second derivative test tell us?
The function has a
local extremum at \(x=a\).The function does not have a local extremum at \(x=a\).It gives no information on whether the function has a local extremum at \(x=a\).