Question

let f(x) = 2x^3-3x^2-36x+14 a. find f(x) b. find the interval on which f is increasing

Answer 1

What do you mean find f(x)? Isn't it given in the question?

Answer 2

no find f'(x)

Answer 3

ok.

f'(x) = 6x^2 -6x -36

This is what I did:
f'(x) = 6[3x^(3-1)] -3[2x^(2-1)] -36[1x^0]
(I'm taking down the exponent and multiplying it by the coefficient in front of it. And I'm also subtracting 1 from the exponent when I do this.)

Understand?

Answer 4

yes, but how do know were f is increasing

Answer 5

For part b:
Critical numbers are x values where f'(x) = 0.
So to see where f(x) is increasing or decreasing we are checking f'(x) to the right and left of each critical number to see if it is less than or greater than zero.
Where it is greater than zero represents increasing.
Where it is less than zero represents decreasing.

So:
0 = 6x^2 -6x - 36

divide both sides by 6:
0 = x^2 -x -6

factor:
0 = (x - 3)(x + 2)

critical numbers:
x = 3, -2

on the left of -2: f'(x) > 0 (increasing)
between -2 and 3: f'(x) < 0 (decreasing)
on the right of 3: f'(x) > 0 (increasing)

Answer 6

Does that help?

Answer 7

Yes that helped a lot thanks

Answer 8

glad to hear

Answer 9

let f(x) =x^3-3x^2+3x+5. a.find the critical points (numbers) of f b. find local maximum and minimum values of f.
So far I only got 2^2-6x+3

Answer 10

f'(x) = 3x^2 - 6x + 3

0 = 3x^2 - 6x +3
0 = 3(x-1)^2

critical number: x = 1

Does that help?