Critical numbers for function - x^3-3x^2+4 on [-2, 4]

[Tazman]

I have a function - x^3-3x^2+4 on interval of [-2,4], what are my critical numbrs:

Can someone check out my work. Would greatly apprec.

Take 1st Deriv: 3x^2-6x
Take 2nd Dervi: 6x-6

3x(x-2) = 0

x = 0,2

 

[stapel]

Why are you using the second derivative to find critical numbers...? :shock:

The function should have been given a name. For convenience, we can give it one:

. . . . .Find the critical numbers of f(x) = -x³ - 3x² + 4 on the interval [-2, 4], if any.

When you took the first derivative, it looks like you lost a "minus" sign; you should have f'(x) = -3x² - 6x (assuming that is the meaning of the "minus" you put in your post). Then set that equal to zero, and solve:

. . . . .-3x² - 6x = 0

. . . . .-3x(x + 2) = 0

. . . . .-3x = 0 or x + 2 = 0

Solve, and then see if either of the solutions is inside the given interval.

Eliz.

 

[Malga1968]

Why are you using the second derivative to find critical numbers...? :shock:

The function should have been given a name. For convenience, we can give it one:

. . . . .Find the critical numbers of f(x) = -x³ - 3x² + 4 on the interval [-2, 4], if any.

When you took the first derivative, it looks like you lost a "minus" sign; you should have f'(x) = -3x² - 6x (assuming that is the meaning of the "minus" you put in your post). Then set that equal to zero, and solve:

. . . . .-3x² - 6x = 0

. . . . .-3x(x + 2) = 0

. . . . .-3x = 0 or x + 2 = 0

Solve, and then see if either of the solutions is inside the given interval.

Eliz.

Sorry bout that - it should not be a minus sign in front of the x^3.

So are my critical numbers x= 0,2.

Also if x = 1 would that mean f(x) is "increasing"?

Thx Eliz..

Taz

 

[o_O]

If the function at x = 1 is increasing, that means f'(x) > 0 (the tangent line at x = 1 has a positive slope). If it is decreasing, then f'(x) < 0 (tangent line at x = 1 has a negative slope).

 

[Malga1968]

If the function at x = 1 is increasing, that means f'(x) > 0 (the tangent line at x = 1 has a positive slope). If it is decreasing, then f'(x) < 0 (tangent line at x = 1 has a negative slope).

Okay. I think I got it, but please confirm:

F(x) would be decreasing as my slope would be a -3.

Just to take another example if when x =5, then f(x) would be increasing and the graph of f(x) would be concave up. Is this right?

Taz

 

[o_O]

Yep, yep, and yep.

 

[Tazman]

Yep, yep, and yep.

THanks!!