Transcript Take it to the Extrema

Take it to the Extrema
Maximum and Minimum Value
Problems
By: Rakesh Biswas
Decreasing and Increasing
Functions
In this study of graphs it is important that
we take a very close look as the when
function are increasing and decreasing.
Knowledge of when a function is
increasing or decreasing will aid our
understanding of what a graph may look
like if we do not have a calculator.
Increasing Function
A function is said to be
increasing if:
f(x1)<f(x2) if x1<x2
This graph shows that if
the above conditions
are satisfied the
function is increasing.
f(x)
f(x2)
f(x1)
x1
x2
Decreasing Function
A function is said to be
decreasing if:
f(x1)>f(x2) if x1<x2
f(x1)
f(x)
f(x2)
This graph shows that if
the above conditions
are satisfied the
function is
decreasing.
x1
x2
First Derivative Test
The First Derivative Test will tell you when
the function is increasing, decreasing or
constant.
First Derivative Tests can also be used to
find the relative extremas.
It will also give you the critical points of the
function.
Used in conjunction with the Second
Derivative Test the graph of a function can
be sketched.
Second Derivative Test
• Second Derivative Tests can be used to
find the concavity of a function.
• It can also be used to find the inflection
points of the function by setting the second
derivative of the function of equal to zero
and solving.
• Once the concavity of the function is found
it can be used along with the results of the
First Derivative Test to sketch the function.
Using the First Derivative
f(x)= x2-3x+8
(a) Find when f is increasing and decreasing.
(b) Find where f has a relative minima and relative maxima.
Solution
f’(x)= 2x-3
0= 2x-3
X=3/2
 f’(1)= 2(1)-3= -1
(a) f(x) is increasing when x>3/2
f(x) when x< 3/2 is negative
f(x) is decreasing when x<3/2
 f(2)= 2(2)-3=1
(b) f(x) has a relative minimum at x=3/2 because the
f’(x) when x>3/2 is positive
the graph f’(x)<0 (negative) when x<3/2 and
f(x)>0 (positive) when x>3/2. f(x) has no relative
extrema.
Using the Second Derivative
Find the intervals of concavity and the
inflection points of g(x)= x4-486x2
g’(x)= 4x3 – 972x
g’’(x)= 12x2 -972
0= 12x2 -972
0= 12(x2 – 81)
X= 9 and x=-9
 g’’(-10)= 12(-10) 2 – 972 = 228
 g’’(0)= 12(0) 2 – 972 = -972
 g’’(10)= 12(10) 2 – 972 = 228
f is concave up on (, 9) and (9, )
f is concave down on (-9,9)
f has inflection points at x=-9 and x=9
because the concavity changes form
concave up to concave down at x=-9
and concave down to concave up at
x=9
(, 9)
f(x)
x=-9
x=9
Visualizing Graphs
The First and Second Derivative tests can
be used to sketch the graphs of a function.
The First Derivative test tells us if the
function is increasing or decreasing and
the Second Derivative test tells us if the
graph is concave up or down.
Both the First and Second Derivative tests
can be used to find the inflection points of
the of the function.
Graph Sketching
In each part sketch a continuous curve y=f(x) with the stated properties.
(a) f(2)=4, f’(2)=0, f’’(x)<0 for all x.
If the first derivative of a function is set equal to zero and the solved for x. The
horizontal tangents can be acquired. In this problem when x=2 the slope of
the tangent is zero (f’(2)=0), and at x=2 the value of y is 4.
The problem also states that the second derivative is less than zero for all values
of x. Since f’’(x)<0 graph of the function is concave down on the interval
(, )
(2,4)
More Graphs
In each part sketch a continuous curve y=f(x) with the stated properties
f(2)=4, f’(2)=0, f’’(x)>0 for x<2, f’’(x)<0 for x>2
 This conditions state that at x=2 there is a horizontal
tangent (f(2)=0).
 It is also stated that when x<2 the function is concave up
f’’(x)>0 and when x>2 the function is concave down f’’(x)<0.
 There is also an inflection point at x=2 because the concavity of the
graph changes for concave up to concave down.
Finding Information form Derivative
Graphs
The graph of the first derivative can be
used to find crucial information about the
function for the original graph.
For example, by using the graph of the
first derivative function, we can find the
relative extremas, points of inflection and
concavity.
Using the First Derivative Graph
The figure below shows the graph of f’, the derivative of f. The domain of f is
[-5,6]
Find:
(a) relative extrema
(b) concavity
f’(x)
(c)
all inflection points
Note: Problem is form
second test.
Relative Extrema
(a)
The relative extrema occur when the graph goes form positive to
negative or form negative to positive. In other words if the graph of the
derivative function is below the x-axis the function has a negative slope in
that interval. If the graph of the derivative function is above the x-axis the
function has a positive slope in that interval. This means that when a
graph goes for positive to negative, there is a relative maximum and
when that graph goes form negative to positive there is a relative
minimum. At x= - 4 the graph goes from negative to positive which means
that there is a relative minimum there. At x=0 there is a relative maximum
because graph goes form positive to negative. And finally, at x=5 there
exists another relative minimum because the graph goes form negative to
positive.
Relative minima at: x=-4 and x=5
Relative maximum at: x=0
Concavity
(b) We can find the concavity of the function by finding the intervals on
which the slope of the derivative function is positive or negative. If
the slope is increasing, the function is concave up and if the slope is
decreasing the function is concave down. Since the slope of the
derivative function is increasing form ( , 5)and (2, ) the
function is concave up. Since the derivative function has a negative
slope in the interval (-2,2) the function is concave down on this
interval.
f is concave up on ( , 5)and (2, )
f is concave down on (-2,2)
Inflection Points
(c) The horizontal tangent(s) of the derivative function is the
inflection point. Another way to find the inflection points
is to find the concavity of the function. When the
concavity of the function changes form positive to
negative or vice versa, an inflection point exists at that
point.
f has inflection points at x=-2 and x-2 because there are
horizontal tangents at these points.
Sketch of Original Function
f(x)
References
Slide 2: Definition 5.1.1
Slide 3: Definition 5.1.1
Slide 4: Problem is form test two
The End