Unit 5 - Analytical Application of Differentiation Flashcards

1
Q

What is the Extreme Value Theorem?

A

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].

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2
Q

What is a critical point?

A

Point with the possibility of being an extrema

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3
Q

How do you determine a critical point?

A

The derivative of the function does not exist or equals zero.

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4
Q

How do you determine a relative extrema?

A

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.

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5
Q

What is the absolute maximum?

A

Point on a function that has the highest y-value

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6
Q

What is the absolute minimum?

A

Point on a function that has the lowest y-value

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7
Q

What is the Mean Value Theorem?

A

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.

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8
Q

What is the equation for the Mean Value Theorem?

A

f’(c) = f(b) - f(a) / b - a

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9
Q

When the slope of a function is positive, then the function is

A

increasing

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10
Q

When the slope of a function is negative, then the function is

A

decreasing

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11
Q

How to find intervals of a function that is increasing or decreasing

A

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

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12
Q

How to justify a function is increasing on an interval

A

Claim that its derivative is positive

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13
Q

How to justify a function is decreasing on an interval

A

Claim that its derivative is negative

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14
Q

First Derivative Test

A

Used to determine where min/max may exist on a function

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15
Q

Minimum exists when

A

the derivative of a function changes sign from negative to positive

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16
Q

Maximum exists when

A

the derivative of a function changes sign from positive to negative

17
Q

“What is the maximum value” is asking for

A

f(c) or the y-value

18
Q

“What is the maximum “ is asking for

A

x=c or the x-value

19
Q

Endpoints

A

Also has the possibility of being a minimum or maximum on an interval

20
Q

Candidates Test

A

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

21
Q

Conavity

A

State of being concave

22
Q

If f is concave up, then

A

f’ is increasing

f’’ > 0

23
Q

If f is concave down, then

A

f’ is decreasing

f’’ < 0

24
Q

How to find intervals of concavity

A

1- Find the second derivative and set it equal to 0

2- Use sign chart to find intervals of concavity

25
Q

Points of Inflection

A

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

26
Q

Two common mistakes

A

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

27
Q

Relative max

A

f’(x) = 0

f”(x) < 0

28
Q

Relative min

A

f’(x) = 0

f”(x) > 0

29
Q

If there is only 1 critical point, and that critical point is an extremum, then

A

it is an absolute extremum

30
Q

The slope of f is the

A

y-value of f’

31
Q

Optimize

A

Make the best or most effective use of a situation or response

32
Q

Strategies for optimization

A

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

33
Q

Implicit relationships follow the same rules as functions when it comes to

A

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