1. Maximum and Minimum Value Problems By: Rakesh Biswas
Take it to the Extrema
Maximum and Minimum Value Problems
By: Rakesh Biswas

2. 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.

3. 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

4. Decreasing Function A function is said to be decreasing if:
f(x1)>f(x2) if x1<x2
This graph shows that if the above conditions are satisfied the function is decreasing.
f(x)
f(x1)
f(x2) x1 x2

5. 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.

6. 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.

7. 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= (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= (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.

8. 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 and
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
f(x)
x=-9
x=9

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.

10. Graph Sketching
In each part sketch a continuous curve y=f(x) with the stated properties.
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)

11. 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.

12. 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.

13. 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:
relative extrema
concavity
all inflection points
f’(x)
Note: Problem is form second test.

14. Relative Extrema
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

15. 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 and 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 and
f is concave down on (-2,2)

16. 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.

17. Sketch of Original Function
f(x)

18. References Slide 2: Definition 5.1.1 Slide 3: Definition 5.1.1
Slide 4: Problem is form test two

19. The End