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M3 Mastery
M3 Mastery
-Melvin Orichi Socana
L'Hopital's Rule
L'Hopital's rule is a concept that can help in solving limits that result indeterminate values.
- As such, it can only be used in questions asking about the value of limits where indeterminate values cannot be avoided.
L'Hopital's Rule involves taking the derivative of a fraction's numerator and denominator. It can be used multiple times if an indeterminate still exists within the limit.
Example Problem
Find:
- The limit as x approaches infinity of the function: x^2/e^-x
Remember that L'Hopital's rule can keep on being applied if the indeterminate value still exists!
Example Problem
Solution
Solution
Solve for the limit inititally of x^2/e^-x as x approache sinfinity yields infinity/infinity which cannot be calculated. Since this is an indeterminate value, apply L'Hopital's rule by taking the derivative of the numerator and denominator:
- limit as x approaches infinity of x^2 / e^-x
- = limit as x approaches infinity of 2x / e^-x
If the value of the limit is calculated again, the same value is obtained, infinity over infinity. Since the indeterminate value still exists, it is fine to use L'Hopital's Rule again:
- limit as x approaches infinity of 2x / e^-x
- = limit as x approaches infinity of 2 / e^-x
Now, when we take the limit by substituting in x, we get the value of 0 as any number except infinity divided by infinity is 0.
Optimization
- Optimization is a common problem of derivatives that always appaers in the AP exam. It pertains with "optimizing," or finding the minimum or maximum values for a desired effect.
- The process of optimizing values usually relies on the First Derivative Test to find the relative extrema of a function.
- Find the critical points of the function
- Determine the values of the derivative near these critical points to solve if there is relative maximum or minimum
Steps for Optimization
- Determine what value needs to be optimized
- Write an equation to be optimized with only 1 independent variable
- Perform the First Derivative Test and solve for x to find Relative Extrema/Critical Points
- Perform a number-line analysis to confirm what type of relative extrema has been located
- Compare local extrema to endpoints to determine if the absolute extrema have been found
- Conclude with correct units
Example Problem
A farmer has 200 feet of fencing to enclose two adjacent rectangular corrals, as shown in the figure. What dimensions should be used so that the enclosed area will be a maximum?
Example Problem
Solution
Solution
In this problem, area is trying to be maximized with specific dimensions:
- A = 2xy
The 200 feet of fencing provides the secondary equation for perimeter which can be used to solve for another variable:
- P = 3x + 4y
- 200 = 3x + 4y
- 200 - 4y = 3x
- x = (200 - 4y) / 3
Now substitute x for the equation for area:
- A = 2((200 - 4y) / 3) * y
- Simplify: A = (400/3)y - (8/3)y^2
Diffrientiate:
- A' = (400/3) - (16/3)y
Solution Continued
Since the First Derivative test must be used, find the critical points by solving for y in the derivative:
- 0 = (400/3) - (16/3)y
- 400/3 = (16/3)y
- y = 25
25 is the only critical point for y, perform a number line analysis to determine the type of relative extrema it is:
Now go back to the equation for x and plug in y to find the final dimension.
- x = (200 - 4(25)/3 = 100/3
- The dimensions the farmer needs are x = 100/3 and y = 25
Since the value of the derivative is changing from positive to negative at the critical point, there is a relative maximum at the critical point