Unit 5 - Analytical Application of Differentiation Flashcards
What is the Extreme Value Theorem?
If a function is continuous over interval [a,b], then f has at least one minimum value and at least one maximum value on [a,b].
What is a critical point?
Point with the possibility of being an extrema
How do you determine a critical point?
The derivative of the function does not exist or equals zero.
How do you determine a relative extrema?
Isolate the suspected point as well as surrounding domain. If this point is higher or lower than the points adjacent to it, then it can be considered an relative extrema.
What is the absolute maximum?
Point on a function that has the highest y-value
What is the absolute minimum?
Point on a function that has the lowest y-value
What is the Mean Value Theorem?
If a function f is continuous and differentiable over the interval [a,b], then there exists a point c within that open interval where the instantaneous ROC equals the average ROC over this interval.
What is the equation for the Mean Value Theorem?
f’(c) = f(b) - f(a) / b - a
When the slope of a function is positive, then the function is
increasing
When the slope of a function is negative, then the function is
decreasing
How to find intervals of a function that is increasing or decreasing
1- Find critical points
2- Between these points, intervals of derivative must be positive or negative
3- Use sign chart to keep track
4- Answer with justification
How to justify a function is increasing on an interval
Claim that its derivative is positive
How to justify a function is decreasing on an interval
Claim that its derivative is negative
First Derivative Test
Used to determine where min/max may exist on a function
Minimum exists when
the derivative of a function changes sign from negative to positive
Maximum exists when
the derivative of a function changes sign from positive to negative
“What is the maximum value” is asking for
f(c) or the y-value
“What is the maximum “ is asking for
x=c or the x-value
Endpoints
Also has the possibility of being a minimum or maximum on an interval
Candidates Test
1 - Find derivative of function and set equal to 0 to find critical points
2 - Substitute critical points & endpoints into original function
3 - Lowest value is considered the abs. min. value
Highest value is considered the abs. max. value
Conavity
State of being concave
If f is concave up, then
f’ is increasing
f’’ > 0
If f is concave down, then
f’ is decreasing
f’’ < 0
How to find intervals of concavity
1- Find the second derivative and set it equal to 0
2- Use sign chart to find intervals of concavity
Points of Inflection
There is a point of inflection of f at x=c if f(c) is defined and f’’ changes sign at x=c
Where graph changes concavity
Two common mistakes
1 - Assuming that f’’ = 0 means there is a point of inflection -> need to check for sign change
2 - Assuming that f’’ is not equal to 0 means there are no points of inflections
Relative max
f’(x) = 0
f”(x) < 0
Relative min
f’(x) = 0
f”(x) > 0
If there is only 1 critical point, and that critical point is an extremum, then
it is an absolute extremum
The slope of f is the
y-value of f’
Optimize
Make the best or most effective use of a situation or response
Strategies for optimization
1 - Draw a picture and identify known and unknown quantities
2 - Write an equation that will be optimized
3 - Write your equation in terms of single variable
4 - Determine desired max or min value with calculus techniques
5 - Determine domain and endpoints of equation to verify if endpoints are max or min
Implicit relationships follow the same rules as functions when it comes to
derivative and second derivatives
if dy/dx = 0 or DNE, then point = critical point
if d22y/dx^2 > 0, then graph = concave up
if d22y/dx^2 < 0, then graph = concave down
