This week all we did was review so I will reiterate on all of chapter three which we have learned so far so that lessons 1-3 are all in one place in case there is a need to print, study, refer to etc...
CHAPTER 3 LESSON 1
Extreme Values are defined as maximums and minimums
Absolute Max is defined as the highest point on an interval
** notice that absolute max is the highest point on a given interval not the highest point on the graph
Absolute min is defined as the lowest point on an interval
**notice that absolute min is the lowest point of a given interval not the lowest point on the graph
Relative Max (also referred to as local max) is defined as any max NOT on an interval
Relative Min (also referred to a local min) is defined as any min NOT on an interval
Critical Numbers are defined as any max or min typically referred to as x=c
When trying to solve problems concerning a critical number, extreme value, max, min, or horizontal tangents take the derivative and set equal to zero.
Example #1:
Find the critical values of f(x)= 9(x^2-3)/x^3
9x^2-27/x^3
Take the derivative
X^3(18x)-[(9x^2-27)(3x^2)]/ x^6
18x^4-[27x^4-81x^2]/x^6
18x^4-27x^4+81x^2/x^6
-9x^4+81x^2/ x^6
9x^2(-x^2+9)/X^6
9(-x^2+9)/x^4
Set equal to zero and solve
((when setting a fraction equal to zero you are really only setting what is in the numerator equal to zero))
9(-x^2+9)=0
-9x^2+81=0
-9x^2=-81
X^2=9
X=+/- 3
Example #2:
Find the extrema of f(x)= 3x^4-4x^3 on the interval [-1,2]
4(3)x^3-3(4)x^2
12x^3-12x^2=0
12x^2(x-1)=0
X=0,1
To determine max or min on an interval plug into original and the biggest result will be the max
and the smallest result will be the min
f(-1)= 3(-1)^4-4(-1)^3=7
f(0)=3(0)^4-4(0)^3=0
f(1)= 3(1)^4 – 4(1)^3= -1 ((smallest so absolute min))
f(2)= 3(2)^4-4(2)^3= 16 ((largest so absolute max))
CHAPTER 3 LESSON 2
Rolle’s Thorem:
Rolle’s Thorem is used to discover whether there is a max or a min on the given interval.
Steps:
Check differentiability
Check for continuity
If the problem is not differentiable or if the problem is not continuous Rolle’s thorem cannot be applied.
If the problem is differentiable and is continuous then you will continue to work the problem by:
Test that the y values match by plugging interval numbers into the original problem
Take the derivative and set equal to zero
Solve for x
**Remember f’(c)= 0 must be read as max or min and therefore you must use rolle’s thorem
Example:
Let F(x)=x^4-2x^2 find all values of c in the interval [-2,2] such that f’(c)=0
Check differentiability
Differentiable? Yes.
Check for continuity
Continuous? Yes
Test that the y values match by plugging interval numbers into the original problem
F(-2)= -2^4-2(-2)^2
16-2(4)
16-8
=8
F(2)= 2^4-2(2)^2
16-2(4)
16-8
=8
Therefore, yes
Take the derivative and set equal to zero
4x^3-4x=0
Solve for x
4x(x^2-1)=0
X=0 X=-1 X=1
Therefore by Rolle’s thorem c= 0, -1, 1
By Rolle’s theorm the function is continuous, differentiable, and f(1)=f(2) thus there must be one max or min on the interval [-2, 2] at f’(x)=0
Mean Value Thorem: The mean value theorem is used to prove that on the interval there must be some x value where the derivative equals the slope between the points of the interval
Steps:
Check Differentiability
Check Continuity
If the problem is not differentiable or if the problem is not continuous the Mean Value theorem cannot be applied.
If the problem is differentiable and is continuous then you will continue to work the problem by:
Check Slope (((( f’ (C)= f(b)-f(a)/b-a))))
Take the derivative and set equal to zero
Solve for x
Example: Given f(x)= 5-(4/x) find all values of c in the open interval (1,4) such that f’(c)= f(4)-f(1)/ 4-1
Check Differentiability
Differentiable? Yes
Check Continuity
Continuous? Yes
Check Slope (((( f’ (C)= f(b)-f(a)/b-a))))
4-1/4-1 = 1
Take the derivative and set equal to slope
-4(-1)x^-2
4x^-2
4/x^2= 1
Solve for x
X=+/- 2
Therefore by the mean value theorem, the function is continuous and differentiable on the interval therefore there is a value c=2 where the derivative = slope on the interval (1,4)
CHAPTER 3 LESSON 3
The first derivative test:
Used to determine whether it is specifically a max or a min on the interval
KEY WORDS TO RECOGNIZE WHEN TO USE THE FIRST DERIVATIVE TEST
Increasing, decreasing, max, min, horizontal tangent, change direction
Steps:
Take derivative and set equal to zero
Solve for x
Set up intervals (-infinity, pt)U(pt, infinity)
Plug in value on the interval into the derivative ((if you get a positive number the function is increasing, If you get a negative number the function is decreasing)) (( A max is increasing then decreasing, A min is decreasing then increasing))
Example:
Find the open intervals on which f(x)= x^3-3/2x^2 is increasing or decreasing
Take derivative and set equal to zero
3x^2-3x=0
Solve for x
3x(x-1)=0
X=0,1
Set up intervals (-infinity, pt)U(pt, infinity)
(negative infinity,0)U(0,1)U(1,infinity)
Plug in value on the interval into the derivative
F’(-1)= 3(-1)^2-3(-1)=6
Positive therefore increasing
F’(1/2)= 3(1/2)^2-3(1/2)= -3/4
Negative therefore decreasing
F’(2)= 3(2)^2-3(2)= 6
Positive therefore increasing
Therefore increasing: (negative infinity, 0) U (1, infinity)
Decreasing: (0,1)
Max: at 0
Min: at 1
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