Sect. 4.2
Sect. 4.3
Sect. 4.4
Sect. 4.6
100

What is the Extreme Value Theorem?

If f is continuous on a closed interval, then f has both a minimum and a maximum on the interval.

100

Definition of critical numbers.

What is f'(c) = 0 or if f(x) is not differentiable at c, c is a critical number?

100

Concavity

All of its tangents lie below the curve on the interval. The slope of the tangent lines are increasing.

What is concave upward?

100

Relating f and f'

f'(x) > 0

means...

f(x) increases

200

Sarah says there is an absolute maximum at an endpoint. May this occur?

What is yes, the absolute max/min of a function may occur at the endpoints.

200

Critical Numbers only exist if the number line test shows that it changes sign at c.

What is false? 

(This is the case for inflection points.)

200

Determine the points of inflection. 

f(x) = x- 4x3

What is x = 0 and x = 2?

300

To find the max and min of f on [a, b]:

Find the critical numbers of f(x) in (a, b)

Evaluate f(x) at each critical number in (a, b).

The least of these values is the minimum. The greatest is the maximum.

What is missing?

Evaluate f(x) at each endpoint in [a, b].

300

If the sign changes from negative to positive at c, it is a relative max.

What is false? 

(- \ to / +)

300

If f''(c) = 0, does it pass the Second Derivative Test?

What is no?

400

Find the absolute maximum and/or minimum of sinx -sinx2, on [0, 2pi].

The absolute max is 1/4 when x = π/6, 5π/6.

The absolute min is –2 when x =3π/2.

400

Find relative max or min of

f(x) = x3 - 9x2 + 24x

Rel max at x = 2

Rel min at x = 4

400

Use the 2nd derivative test to find the local max and/or min for ex - 5x.

What is a local minimum at x = ln5?

500

Rayla was given a problem. She believes she made a mistake somewhere but is unsure. Did she make a mistake? If so what is it?

Find the absolute extrema, if they exist, for the function

f(x) = x^2/(x-1) on [-2,3]

f’(x) = {2x(x-1) - x^2}/(x-1)^2

f’(x) = (x^2 - 2x)/(x-1)^2

f’(x) = {x(x-2)}/(x-1)^2

Critical Numbers: x = 1, x = 0, and x = 2

x = 1 can not be a critical number as it is not in the domain

500

Relative max is (-1, cube root of 2)

Relative min is (1, -cube root of 2)

500

Use the 2nd derivative test to find the local max and/or min for f(x) = 10x6 - 24x5 + 15x4

What is a local min at x = 0?