APPLICATIONS OF DERIVATIVES
In this chapter we give some applications of derivatives.
We say that f has an absolute
maximum value Mon
an interval I if
for all x in I.
We say that f has an absolute
minimum value mon
an interval I if
for all x in I.
The Maximum-Minimum
Theorem for Continuous Functions:
If f is continuous at each point of a closed interval,
then f assumes both an absolute
maximum value M and an absolute
minimumvalue m somewhere
in I.
Local Extremes Values:
We say that f has an local
maximum value atan
interior point c of its domain I if
for all x in some open interval containing c.
We say that f has an local
minimum value atan
interior point c of its domain if
for all x in some open interval containing c.
The First Derivative
Theorem for Local Extreme Values:
If f has a local maximum or minimum value at an interior
point c of its domain then
provided 
is defined.
Finding Extreme Values:
The place where a function f can possible have extreme values;
3.endpoints of the domain of f.
Exercises: Find the
absolute maximum and minimum values of each function. Verify our answers
graphically.
So we need to check the values of f at x = 0 and the
endpoints x = -2 and x = 3.
f(0) = -4
f(-2) = 0
f(3) = 5
So the function has an absolute maximum value of 5
at x = 3 and an absolute minimum value of -4 at x =0.
which zero when x = 0.
So we need to check the values of f at x = 0 and the
endpoints x = -2 and x = 3.
f(0) = 3
f(3) = 0
So the function has an absolute maximum value of 3
at x = 0 and an absolute minimum value of 0 at x =3.
Answer the following question:
What are the critical points of f?
On what intervals f is increasing or decreasing?
At what points, if any, does f assume local maximum
and local minimum values?
Verify your results graphically.
So f assumes local maximum at x = -1 and local minimum
at x = 2.
Now follow #3.
Critical points of f are : x = 1 and x = 5.
f(x) has local maximum at x =1 and local minimum at
x = 5.
f has no absolute maximum or minimum.
It is 0 when x = 1.
F is increasing for x > 1 and decreasing for x <
1.
So f has a local minimum at x = 1. It has also absolute
minimum at x =1 but no absolute maximum.
Concavity:
The graph of a differentiable function y = f(x) is concave up on an interval if
and concave down
on an interval if
The Second Derivative
Test for Concavity:
Let y = f(x) be differentiable such that its
first two derivatives exist on an interval I.
Points
of Inflection:
A point on the graph of
y = f(x) where it has a tangent and where the concavity changes is called
a point of inflection.
Note:The
second derivative of f is 0 at a point of inflection.
Examples: Verify your answers graphically.
So x = 0 is a point
of inflection.
f is concave down at x=
0
f is concave up at these
points.
The height of an object t seconds after it is launched
vertically into the air is given by
a.Sketch the graph of s(t)
b.Find an equation for its velocity.
c.What the initial velocity of the object?
d.For which time the object is rising, falling?
e.What
is the maximum height attained by the object?
Graph
c. initial velocity = 128 ft/sec (when t = 0)
d. The object stops rising when v = 0 that is 128
– 32 t = 0. Then t = 4 sec.
It hits the ground when s = 0. That is when t = 0 or 8 sec.
So the time to rise = time to fall = 4 sec.
e. maximum height = 128(4) – 16(4)^2 =
256 ft. when t = 4 sec
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The figures on the right has a graph of
y = f(x)and its derivative. Identify each graph. |
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Applications: Check your answers graphically.
1.The
perimeter of a rectangular garden is 216 sq.ft. Find its dimensions
so that its area is maximum.
Let its length and width be x ft. and y ft.
Perimeter 2x + 2y =216
Area = xy
will give us x = 54 and y = 54.
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2. Find the dimensions of the right circular cylinder
of largest volume that can be inscribed in a right circular cone with base
radius 12 inches and height 36 inches.
Solution:
From the two similar triangles on the right , we have
h = 36 - 3r
will give us r = 0 and r = 8.
Maximum volume when r = 8 and h = 12
Find the area of the largest rectangle with base on the x-axis and upper vertices on the parabola
From the diagram on the right, area of the rectangle
Now by letting its derivative = 0, find x.
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Inscribing a rectangle in a semicircle of radius
4 inches with center at (0,0):
Its area
From example 3, its derivative is 0 when
and so the maximum area will be for
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5. For what values of a the function
has a
a.local minimum at x =2
b.a point of inflection at x =1.
c.a
local maximum for some a
when x = 2 gives us a = 32.
for all x.
So f(x) has a local minimum when a = 32 and
does have a local maximum.
Substituting the second derivative = 0 with x = 1,we
get a point of inflection when a = -2.
Linearization of a Function:
Linearization of a Function:
If f is a differentiable
function at x = a, then the linearization of f at x = a is given
by
This means f(x) can be approximated by the linear function L(x) near a.
Examples:
For x = 1, L(x) = ¼ + 3/2 = 1.75 with an error = f(1) – 1.75 = sqrt(3) – 1.75 = - 0.017949
For x = 3, L(x) = 2.25, with an error of sqrt(5) – 2.25 = -0.013932
For x = 1.5, L(x) = 1.875, with an error of sqrt(3.5) – 1.875 = -0.004171
For x = 2.5, L(x) = 2.125, with an error of sqrt(4.5) – 2.125 = -0.003679
#2
Find a linearization
of at x = 2 of:
L(x) = 3 + 8(x – 2) = 8x –13
#3 Approximate f(x) = sin x near x = 0.
f(0) = 0
L(x) = 0 + 1(x – 0)
= x
#4Approximate
f(x) = sec x near x = pi
L(x) = -1 + 0(x – pi)
= -1
#5
near x = 0.
f(0) = 1
