calculus help! Wut the????

Question

Consider the function f(x) = 12 x5 + 45 x4 −200 x3 + 3. For this function there are four important intervals: (−∞, A], [A,B],[B,C], and [C,∞) where A, B, and C are the critical numbers.
Find A__?
and B__?
and C__?
At each critical number A, B, and C does f(x) have a relative min, a relative max, or neither? Type in your answer as RMIN, RMAX, or NEITHER.
At A__?
At B__?
At C__?

Answer 1

Solve f'(x) = 0:
60x^4 + 180x^3 - 600x^2 = 0
60x^2(x^2 + 3x - 10) = 0
60x^2(x+5)(x-2) = 0

Critical #s: -5, 0, 2

Now test the second derivative at each point. If the second derivative is positive, it is a relative minimum. If negative, it is a relative max. O/W neither.
60(4x^3 + 9x^2 - 20x)

Answer 2

Take the first derivative and set equal to zero, then solve:

60x^4 + 180x^3 - 600x^2 = 0

60x^2(x^2 + 3x - 10) = 0

60x^2(x + 5)(x - 2) = 0

In order for the equation to be zero, x = 0, -5, or 2.

A = -5
B = 0
C = 2

To find the answer to the second part, solve f(x) for

A) x = -6, -5, -4
B) x = -1, 0, 1
C) x = 1, 2, 3

And see what the values are. If they are the critical numbers are higher than the other two, then you have an RMAX, if the critical numbers are lower, then you have an RMIN, if they are in the middle, then you hvae NEITHER.

Good luck!

Answer 3

The derivative of the function is:

60x^4 + 180x^3 - 600x^2

Set this equal to zero

(60x^2) (x^2 + 3x - 10) = 0

x=0 is one of the critical numbers

Factor x^2 + 3x - 10

(x+5)(x-2) = 0

x = -5 and x = 2

A = -5
B = 0
C = 2

Find the second derivative

240x^3 + 540x^2 - 1200x

Plug A, B and C into this.

If the result is negative, then it is RMAX
If the result is positive, then it is RMIN
If the result is zero, then it is NEITHER