Unit 5

Critical Number

If f'(c) = 0 or does not exist, and c is in the domain of f, then c is a critical number. (Derivative is 0 or undefined)

Mean Value Theorem

If f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), there exists a number c in (a, b) such that AROC = IROC

When f'(x) is above the x-axis, f(x) is...

Increasing

f''(x) = 0 and f'(x) > 0

Increasing and POI

f''(x) < 0 and f'(x) > 0

Increasing and concave down

f''(x) > 0 and f'(x) > 0

Increasing and concave up

Rolle's Theorem

Let f be continuous on [a,b] and differentiable on (a,b). if f(a) = f(b), or the y-values are the *same*, then there is at least one number c on (a,b) such that f'(c) = 0 (If the slope of the secant is 0, the derivative must = 0 somewhere in the interval).

How to find absolute extrema, provided that EVT exists, using *Candidates Test*

1. Find derivative 2. Set equal to 0 - these will be your critical values 3. Perform *Candidates Test* where you test values below and above the critical number 4. Wherever it changes sign is where extrema occur

How to apply intermediate value theorem when given an interval [a, b] such that f(c) = #

1. Find f(a) and f(b). These will be the endpoints of the interval -> f(a) < # < f(b) 2. Determine if the number fits between the two. If so, by MVT there exists a value x = c such that f(c) = #

Easier method of second derivative test without sign charts

1. Find first derivative. Set it equal to 0 and find critical values. 2. Find second derivative. Plug in the critical values from step 1 for x. If positive = concave up; if negative, concave down.

How to find intervals of increasing/decreasing

1. Find the critical values of f, including discontinuities, in the interval (a, b). 2. Create a number line chart using these critical numbers. 3. Choose x-values in between the critical numbers to test the sign of f′ on your sign chart. Your sign chart is NOT A VALID JUSTIFICATION. You MUST write a clear and concise statement communicating your mathematical language related to sign changes of the derivative. DO NOT USE PRONOUNS! For example, "The function g is increasing on the interval −(2, 5) since g′(x) > 0."

How to analyze and sketch f(x)

1. Find x and y intercepts 2. Find first derivative and determine which intervals f(x) is increasing and decreasing 3, Determine relative extrema 3. Find second derivative and determine which intervals f(x) is concave up and concave down 4. Identify any POIs 5. Sketch the graph

Is there a value, a, that guarantees a value c, a < c < b, at which f'(c) = 1/3, with which MVT can be applied on [a, 6]? (FROM GRAPH, 1/7 WARM UP)

1. First, see if f is continuous and differentiable on the interval that is given 2. If it is, MVT can be applied 3. Using MVT, make f'(c) = 1/3 and plug in values from the interval [a , 6] into the AROC formula. Set f'(c) = AROC. 4. Solve for a and f(a) so that you get 1/3. 5. Yes, there is a value a = 3 that guarantees a value c, a < c < b, at which f'(c) = 1/3. The Mean Value Theorem can be applied because it was given that f(x) is differentiable on (0, 6) and continuous on [0, 6].

Optimization problems over open intervals

1. Identify two equations 2. Solve for one of the variables, preferably y. 3. Plug y into the original equation so that there are only x's in the equation, then simplify. 4. Find f'(x) and set equal to 0. This will give you x. 5. Plug x into the original equation to find y.

Optimization: area and perimeter

1. Identify what the equations are for both area and perimeter 2. If you already know the exact area, set that equal to the equation (vice versa if you know the perimeter). Then, solve for y 3. Plug y into the equation for perimeter (or area, depending on what was done in step 2). Simplify 4. Find f'(x) and set it equal to 0. Solve for x, keeping in the domain 5. Plug x into original y equation to find y

How to solve problem involving MVT

1. State whether or not the function is differentiable and continuous 2. Find the derivative 3. Set AROC = IROC 4. Solve for x 5. Therefore c = x satisfied MVT on the interval given

How to answer: Find the number c that satisfies the conclusion of Rolle's Theorem for f(x)

1. State whether or not the function is differentiable and continuous 2. Find the derivative and set it equal to zero. The value(s) will be important later. 3. Show that the y-values are the same and equal zero 4. Set IROC = to AROC. If one side equals zero, then the x value from step 2 satisfies Rolle's Theorem.

How to do second derivative test

1. Take first derivative 2. Set equal to zero and solve - these will be critical numbers 3. Take second derivative 4. Plug in critical numbers for x and solve - Note whether the outcome is negative or positive, which will help determine concavity

How to do first derivative test

1. Take first derivative 2. Set equal to zero and solve, these are your critical values 3. Use a sign chart where f' is on top and f is on bottom. Test values by plugging in numbers above and below the critical number(s) into the first derivative equation 4. Where f'(x) changes sign, extrema occurs 5. Negative = decreasing; Positive = increasing

How to find concavity

1. Take first derivative 2. Take second derivative 3. Set equal to zero and solve, these are possible points of inflection (where f''(x) changes sign) 4. Use a sign chart where f'' is on top and f is on bottom. Test values by plugging in number above and below the critical number. 5. Negative = concave down; Positive = concave up

On the graph of f'(x), wherever there is an x-intercept, there is...

An extrema. If that point goes from above the x-axis to below the axis (positive to negative), there's a max on f(x) If that point goes from below the x-axis to above the axis (negative to positive), there's a min on f(x)

Inflection points can and cannot be at...

Can: - On a continuous function - Sharp turn Cannot: - At a vertical tangent line - Cusp

If f(x) is a continuous function, then the derivative can only change signs at a ________. And, if f(x) is a discontinuous function, then the derivative could change signs at a _________ or at a _________.

Critical value Critical value or at a discontinuity

When f'(x) is below the x-axis, f(x) is

Decreasing

f''(x) = 0 and f'(x) < 0

Decreasing and POI

f''(x) < 0 and f'(x) < 0

Decreasing and concave down

f''(x) > 0 and f'(x) < 0

Decreasing and concave up

Monotonic

First derivative does not change sign

f''(x) = 0 and f'(x) = 0

Function is "smooth level" and a possible POI

Extreme Value Theorem

If f is continuous on [a,b] then f has an absolute maximum and an absolute minimum on [a,b]. The global extrema occur at critical points in the interval or at endpoints of the interval. *Even if the function is not continuous at a point, there may still be a maximum or minimum

Relative/Local Extrema can only occur at a critical value on an ________ interval. (Absolute/Global) Extrema must occur at a critical number OR at an ________ of an interval.

OPEN ENDPOINT

Extrema of f'(x) are possible _______ for f(x)

POIs

f''(x) < 0 and f'(x) = 0

Rel. max and concave down

f''(x) > 0 and f'(x) = 0

Rel. min and concave up

Optimization: shoreline problems and optimizing time

Use pythagorean theorem

Second Derivative Test

Used to determine on what intervals a function is concave up/concave down and the points of inflections.

First Derivative Test

Used to determine where a function's graph has a min/max and is increasing or decreasing.

Optimization: open box and optimizing volume

V(x), or volume, = ldw length and width will most likely be the dimensions - 2x, while height = x

Where there is extrema on the graph of f, there are _________ on the graph of f'

X-intercepts

Point of inflection

f''^(x) = (0 or undefined) and f''(x) changes sign

Derivative of (0.93)^t

ln(0.93)0.93^t

Absolute Max

the largest value of the function, over the entire domain of a function

Absolute Min

the smallest value of the function, over the entire domain of a function

When is f concave down?

when f''(x) < 0

When is f concave up?

when f''(x) > 0