Calculus Homework Help

1. For -3 ≤ x ≤ 4, Evaluate f at its critical points and at the endpoints. Identify global maximum and minimum of f

2. Determine the Open intervals on which the graph is concave upward or concave down.Then, find any inflection points of f..........f(x)=x^3(x-4)

Assuming that both questions use the same function, f(x)=(x-4)(x^3)=x^4-4x^3,

then:

1.first derivative of f=4x^3-12x^2, set this equal to 0, then f ' = 4x^2(x-3). critical points are at: x=0 and 3. The global maximums is at x=(-3). The global minimum is at x=3.
f(-3)=189, f(0)=0. f(3)=-27. f(4)=0.

2. second derivative of f=12x^2-24x, set this equal to 0, and divide out common constants, then x(x-2)=0. x=0&2. so x=0&2 are the x-ordinates of the points of inflection. (0,2) is concave down.(-<><>,0) U (2,<><>) is concave up.

1. I'm assuming that f is the same function you gave in problem 2.

f(x) = x^3(x - 4) = x^4 - 4x^3

f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3); the critical points are at 0 and +3.

f(0) = 0; f(3) = -27.......at the endpoints f(-3) = +189, and f(4) = 0

As f(x) is a fourth degree polynomial, there is no global maximum; global minimum occurs at (3, -27)

2. Now we need second derivative: f''(x) = 12x^2 - 24x = 12x(x - 2); this is zero at x = 0 and 2..........now look at signs of f''. f''(-1) = +36; f''(1) = - 12; f''(3) = + 36.

f''(x) changes sign at 0 and 2, so those are the inflection points.

f(x) in concave up on (- inf, 0) and (2, + inf) and concave down on (0, 2)

1. What is f for this question?

2. f(x) = x^3 (x - 4) = x^4 - 4x^3

f'(x) = 4x^3 - 12x^2 = 0
4x^2(x - 3) = 0
x = 0 or x = 3

f"(x) = 12x^2 - 24x = 0
12x(x - 2) = 0
x = 0 or x = 2

The graph is concave upward on (-∞, 0) and (2, ∞); it is concave downward on (0, 2).

Inflection points are (0, 0) and (2, -16).