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Announcements
Topics:
- finish section 5.4 (derivatives of inverse trig functions), work
on sections 5.6 (second derivative), 5.7 (taylor polynomials),
6.1 (extreme values) and 6.4 (l’Hopital’s rule)
* Read these sections and study solved examples in your
textbook!
Work On:
- Practice problems from the textbook and assignments from
the coursepack as assigned on the course web page (under
the link “SCHEDULE + HOMEWORK”)
Derivatives of Inverse Trig Functions
d
1
(arcsin x) =
dx
1- x 2
d
1
(arctan x) =
2
dx
1+ x
Example 1:
Differentiate.
(a) f (x) = arctan(5x) + 5arctan x
(b) g(x) = ln(arcsin(5x 3 +1))
Example 2:
Prove d
1
(arctan x) =
.
2
dx
1+ x
The Second Derivative
The derivative of the derivative is called the
second derivative.
d2 f
the second derivative of f = f "(x) = 2
dx
The Second Derivative
f” provides information
about f’ and f:
When f’’ is positive, f’ is
increasing, i.e., the rate at
which f is changing is
increasing.
When f’’ is positive, the
slopes of the tangents to the
graph of f are increasing and
the graph of f is concave up.
The Second Derivative
f” provides information
about f’ and f:
When f’’ is negative, f’ is
decreasing, i.e., the rate at
which f is changing is
decreasing.
When f’’ is negative, the
slopes of the tangents to the
graph of f are decreasing and
the graph of f is concave
down.
The Second Derivative
When the graph of f
changes concavity at a
point in the domain of f,
this point is called an
inflection point.
Note:
At an inflection point,
f”=0 or f’’ D.N.E.
The Second Derivative
Example:
Gamma Distribution
g(x) = xe-x ,
x ³0
Find the first and second
derivatives of g(x) and use
them to sketch a graph.
Approximating Functions with
Polynomials
Polynomials have many nice properties and are
generally very easy to work with.
For this reason, it is often useful to approximate
more complicated functions with polynomials in
order to simplify calculations.
Linear functions are the simplest polynomials
and can be used to represent a more
complicated function in many situations.
Linear Approximations
The secant line connecting points (a, f(a)) and (b, f(b))
on the graph of f(x) provides a decent linear
approximation to f(x) for x-values in the interval [a, b].
The tangent line to the graph of f(x) at (a, f(a))
provides the best linear (straight line) approximation
to f(x) near x=a.
Linear (or tangent line) approximation to f(x) around
x=a:
L(x) = f (a) + f '(a)(x - a)
Linear Approximations
Example:
Let f (x) = x .
Find the secant line
approximation to f(x) for
x-values between 1 and 4.
Use this to approximate
both 2 and 3 .
Compare to the actual
values.
Using Technology:
2 »1.41421356237
3 »1.73205080757
Linear Approximations
Example:
Let f (x) = x .
Find the tangent line
approximation to f(x) at
x=1.
Use this to approximate
both 2 and 3 .
Compare to the actual
values.
Using Technology:
2 »1.41421356237
3 »1.73205080757
Quadratic Approximations
We can obtain a more accurate approximation
by using polynomials of higher degrees.
The tangent line (linear)
approximation matches the
value and the slope of the
function at x=1.
The quadratic approximation
matches the value, the slope,
and the curvature (concavity)
of the function at x=1.
L(x) =1+ 12 (x -1)
f (x) = x
T2 (x) =1+ 12 (x -1) - 18 (x -1)2
Quadratic Approximations
To find the quadratic approximation to a
function f (x) at the base point a, we match the
value, the first derivative, and the second
derivative at the point a.
Quadratic approximation to f (x) around x=a:
f ''(a)
T2 (x) = f (a) + f '(a)(x - a) +
(x - a) 2
2
because the
degree is 2
Quadratic Approximations
Example:
1
.
Let f (x) =
1+ x
Find the quadratic
approximation to f(x)
around x=1.
•
f (x) =
1
x +1
The Taylor Polynomial
Suppose the first n derivatives of the function f
are defined at x=a.
Then the Taylor polynomial of degree n
matching the values of the first n derivatives is
nth derivative
The Taylor Polynomial
Example:
Find the 1st, 3rd, and 5th degree Taylor
polynomials for f (x) = sin x near x=0.
The Taylor Polynomial
P1(x) = x
P5 (x) = 5!1 x 5 - 3!1 x 3 + 1!1 x
f (x) = sin x
P3 (x) = - 3!1 x 3 + 1!1 x
Maximum and Minimum Values
f (c) is a global (absolute) maximum of f if
f (c) ³ f (x) for all x in the domain of f .
f (c) is a local (relative) maximum of f if
f (c) ³ f (x) for all x in some interval around c.
Maximum and Minimum Values
f (c) is a global (absolute) minimum of f if
f (c) £ f (x) for all x in the domain of f .
f (c) is a local (relative) minimum of f if
f (c) £ f (x) for all x in some interval around c.
Extrema
Identify the labeled points as local maxima/minima,
global maxima/minima, or none of these.
Extrema
Identify the labeled points as local maxima/minima,
global maxima/minima, or none of these.
local max
global max
local
max
nothing special
local
min
global min
Extreme Values
Notice:
Extreme values occur at either a critical number of f
or at an endpoint of the domain.
(However, not all critical numbers and endpoints
correspond to an extreme value.)
Also note:
By definition, relative extreme values do not occur
at endpoints.
Finding Local Maxima and Minima
(First Derivative Test)
Assume that f is continuous at c, where c is a critical
number of f.
If f’ changes from + to - at x=c, then f changes from
increasing to decreasing at x=c and f(c) is a local maximum
value.
If f’ changes from - to + at x=c, then f changes from
decreasing to increasing at x=c and f(c) is a local minimum
value.
If f’ does not change sign at x=c, then f doesn’t have an
extreme value at x=c.
Finding Local Maxima and Minima
(First Derivative Test)
Example:
Use the first derivative test to find the local
extrema of the following functions.
(a) f (x) = x ln x
x
(b) g(x) =
2
1+ x
Finding Local Maxima and Minima
(Second Derivative Test)
Assume that f’’ is continuous near c and f’(c)=0.
If f’’(c)>0 then the graph of f is concave up at x=c and f(c)
is a local minimum value.
If f’’(c)<0 then the graph of f is concave down at x=c and
f(c) is a local maximum value.
If f’’(c)=0 or f”(c) D.N.E. then the second derivative test
doesn’t apply and you have to use the other method.
Application
Assignment 43, #1 (modified):
Consider the function f (t) = Ate - bt , where A, b > 0.
(a) Find the critical number of f.
Application
Assignment 43, #1 (modified):
(b) Use the second derivative test to determine
if the critical number in part (a) corresponds to a
local maximum, local minimum, or neither.
Application
Assignment 43, #1 (modified):
(c) Determine the values of A and b such that f
describes the graph given below.
Extreme Value Theorem
If f (x) is continuous for all x Î [a, b] , then there are
points c1, c 2 Î [a, b] such that f (c1) is the global
minimum and f (c 2 ) is the global maximum
of f (x) on [a, b].
In words:
If a function is continuous on a closed, finite
interval, then it has a global maximum and a global
minimum on that interval.
Finding Absolute Extreme Values on a
Closed Interval [a,b]
1. Find all critical numbers in the interval.
2. Make a table of values.
The largest value of f(x) is the absolute maximum and
the smallest value is the absolute minimum.
Finding Absolute Extreme Values on a
Closed Interval [a,b]
Example:
1
2
3
Find the absolute extrema of g(x) = x (x - 2) on [-1, 1].
L’Hopital’s Rule
Another application of derivatives is to help
evaluate limits of the form
where either
or
Idea:
Instead of comparing the functions f(x) and g(x),
compare their derivatives (rates) f’(x) and g’(x).
L’Hopital’s Rule
Suppose that f and g are differentiable functions
such that
is an indeterminate form of type 00 or ¥¥ .
If g¢(x) ¹ 0 near a (could be 0 at a) then
L’Hopital’s Rule
Evaluate the following limits using L’Hopital’s
Rule, if it applies.
(a)
(b)
(c)
(d)
L’Hopital’s Rule
Evaluate the following limits using L’Hopital’s
Rule, if it applies.
(a)
(b)