Chapter 3 Application of Derivatives
3.1 Extreme Values of Functions
Absolute maxima or minima are also referred to as global maxima or minima.
Examples
Extreme Value Theorem
The requirements in Theorems 1 that the interval be closed and finite, and
That the function be continuous, are key ingredients.
Local (Relative) Extreme Values
Finding Extrema
Critical Point
Theorem 2 says that a function’s first derivative is always zero at an
interior point where the function has a local extreme value and the
derivative is defined.
Hence the only places where a function f can possibly have an extreme
value (local or global) are
1. Interior points where f’=0
2. Interior points where f’ is undefined,
3. Endpoints of the domain of f.
How to Find the Absolute Extrema
Thus the only domain points where a function can assume extreme
values are critical points and endpoints.
Example
Example. Find the absolute maximum and minimum values of
f(x)=10x(2-lnx) on the interval [1, ex].
Solution. We evaluate the function at the critical points and the
endpoints and take the largest and the smallest of the resulting
values.
The first derivative is
f’(x)=10(2-lnx)-10x(1/x)=10(1-lnx).
Let f’(x)=0, we have x=e. Then
Critical point value: f(e)=10e
Endpoint values: f(1)=20, and f(e2)=0.
So the function’s absolute maximum value
is10e at x=e. The absolute minimum value
is 0 and occurs at the right endpoint x=e2.
3.2 The Mean Value Theorem
Mathematical Consequences
3.3 Monotonic Functions and the First Derivative Test
A function that is increasing or decreasing on an interval is said to be
monotonic on the interval.
Graphs of functions
Each tangent line
Has positive slope.
Each tangent line
Has negative slope.
Each tangent line
Has zero slope.
Theorem
Example
Example: Find the critical points of f(x)=x3 -12x-5 and identify the intervals
on which f is increasing and on which f is decreasing.
Solution: The function f is everywhere continuous and differentiable. The
first derivative f ’(x)=3x2-12=3(x2-4)=3(x+2)(x-2) is zero at x=-2 and x=2.
These critical points subdivide the domain of f into intervals (- , -2), (-2, 2)
and (2, ) on which f ‘ is either positive or negative. We determine the sign
Of f ‘ by evaluating f at a convenient point in each subinterval.
Interval
f ‘ evaluated
Sign of f ‘
Behavior of f
- <x<-2
-2<x<2
2<x<
f ’(-3) = 15
f ‘(0)=12
f ‘ (3) =15
+
increasing
decreasing
+
increasing
First Derivative Test for Local Extrema
Example
Find the critical points of f(x)=(x2-3)ex. Identify the intervals on which f is
increasing and decreasing. Find the function’s local and absolute
extreme values.
Solution.
f ’(x)=(x2+2x-3)ex. Since ex is never zero, f ‘(x) is zero iff x2+2x-3=0.
That is, x=-3 and x=1.
We can see that there is a local maximum about 0.299 at x=-3 and a
local minimum about -5.437 at x=1. The local minimum value is also
an absolute minimum, but there is no absolute maximum. The
function increases on (-, -3) and (1, ) and deceases on (-3, 1).
3.4 Concavity and Curve Sketching
Two ways to characterize the concavity of a differentiable function f on an
open interval:
• f is concave up on an open interval if its tangent lines have increasing
slopes on that interval and is concave down if they have decreasing slopes.
• f is concave up on an open interval if its graph lies above its tangent lines
on that interval and is concave down if it lies below its tangent lines
Concavity and the Second Derivative Test for Concavity
If y=f(x) is twice-differentiable, we will use f’’ and y’’ interchangeable
When denoting the second derivative.
Example
3
Example: Find the intervals on which f ( x)  x 12 is concave up and
the intervals on which it is concave down.
Solution:
f '( x)  3x 2
f ''( x)  6 x
f ''( x)  0 on the interval (0, ),
f ''( x)  0 on the interval (,0).
Thus f(x) is concave up on the interval (0, ), and concave
down on the interval (,0).
Inflection Points
Second Derivative Test for Local Extrema
This test requires us to know f’’ only at c itself and not in an interval about c.
This makes the test easy to apply.
However, this test is inconclusive if f’’=0 or if ‘’ does not exist at x=c.
When this happens, use the First derivative Test for local extreme values.
Strategy for Graphing y=f(x)
Graphical Behavior of Functions from Derivatives
3.5 Parametrizations of Plane Curves
Parametric Formula for dy/dx
Example. Find dy/dx as a function of t if x = t - t2, y = t - t3.
Solution.
dy dy / dt 1  3t 2


dx dx / dt 1  2t
3.7 Indeterminate Forms and L’Hopital’s Rule
How to Use L’Hopital’ Rule
Example
sin 3 x
x 0
5x
Example: Find lim
Solution: The limit is a indeterminate form of type 0/0.
Applying L’Hopital’s rule yields
d
[sin 3x]
sin 3x
3cos3x 3
lim
 lim dx
 lim

x0
x

0
x

0
d
5x
5
5
[5 x]
dx
Example
e3 x  1
Example: Find lim 3
x 0
x
Solution: The limit is a indeterminate form of type 0/0.
Applying L’Hopital’s rule yields
d 3x
[e  1]
3x
e 1
3
e
lim 3  lim dx
 lim 2  
x0
x0
x0 3 x
d 3
x
[x ]
dx
3x
Indeterminate Forms of Type /
Example
x2
Example: Find lim 5 x
x e
Solution: The limit is a indeterminate form of type
Applying L’Hopital’s rule yields
x2
2x
2
lim 5 x  lim 5 x  lim
0
x e
x 5e
x 25e5 x
In fact, we can use LHopital’s rule to show that
xn
ex
lim
 0 and lim n  
x e x
x x
 / 
Example
Example: Find lim
x 
ln x
x
Solution: The limit is a indeterminate form of type
 / 
Applying L’Hopital’s rule yields
1
ln x
1
x
lim
 lim  lim  0
x x
x 1
x x
Similar methods can be used to find the limit of f(x)/g(x) is an
Indeterminate form of the types: 0 ,   ,00 , 0 ,1
3.8 Newton’s Method
Newton’s method is a technique to approximate the solution to an
equation f(x)=0. Essentially it uses tangent lines in place of the
graph of y=f(x) near the points where f is zero.
(A value of x where f is zero is a root of the function
f and a solution of the equation f(x)=0.)
To find a root r of the equation f(x)=0,
• select an initial approximation x1. If f(x1)=0, then r=x1. Otherwise,
use the root of the tangent line to the graph of f at x1 to approximate
r. Call this intercept x2 .
x2  x1 
f ( x1 )
f '( x1 )
• We can now treat x2 in the same way we did x1. If f(x2 )=0, then r=
x2 . Otherwise, we construct the tangent line to the graph of f at x2,
and take x3 to be the x-intercept of the tangent line.
f ( x2 )
x3  x2 
f '( x2 )
Continuing in this way, we can generate a succession of values x1,x2,,
x3,,,x4…that will usually approach r.
Example
Use Newton’s Method to approximate the real solutions of x3-x-1=0
Solution: Let f(x)=x3-x-1. Since f(1)=-1 and f(2)=5, weh know by the
Intermediate Value Theorem, there is a root in the interval (1, 2).
We apply Newton’s method to f with the starting values x0 =1. The
result are displayed in the following table and figure.
At n=5, we come to the result x6=x5=1.324717957. When xn+1=xn,
Equation (1) shows that f(xn)=0. We have found a folution of f(x)=0
to nine decimals.
3.9 Hyperbolic Functions
Identities and Derivatives for Hyperbolic Functions
Inverses Hyperbolic Sine, Cosine, and Secant of x
Inverse Hyperbolic Tangent, Cotangent, and Cosecant of x
Identities and Derivatives of Inverse Hyperbolic Functions