find the set of values of x for which f(x) is increasing?

Question

f(x)≡ 7+24x+3x²-x³

Answer 1

You could find its first derivative and see when it is greater than zero, indicating a positive slope and therefore increasing - assuming you are in calculus.

Answer 2

First off, you need to take the derivative, make f'(x) = 0, and find out the interval where it is positive and negative on f'(x). In this case:

f'(x) = 24 + 6x + 3x^2 . Making it equal to zero and rearranging the polynomial, we get the quadratic:

3x^2 + 6x + 24 = 0. This is now a simple method of getting the roots. We can actually divide that entire equation by 3, to get

x^2 + 2x + 8 = 0
(x+1)^2 + 7 = 0
(x+1)^2 = -7

This has no roots, so f(x) is either always increasing or always decreasing. What you have to do at this point is test *any* value for x^2 + 2x + 8 (let's use zero). you get 0^2 + 2(0) + 8 = 8, a positive number, so f(x) is always increasing.

Therefore, the set of values x for which f(x) is increasing is:

The set of all values x such that x is an element of all the real numbers.

Answer 3

The derivative of f(x) shall be defined as f'(x), then
f'(x) = 0 + 24 + 6x - 3x^2
Set f'(x) = 0, where the slope is zero, there is a min or max
Solve for x when f'(x) = 0, and see what happens as you approach these two (quadratic) values of x.
Using the quadratic formula, I got f'(x) = 0 when x=-2, 4
Now, make a table:
what happens to f(x) when x is in (-infinity,-2), then (-2,4), then
(4,+infinity). You will have your answer.

Answer 4

This souynds like I question I had in precalculus
Being in precalc, I did not know my derivatives yet, so graph it and on the parts of the graph that there is a positive slope it is increasing on the parts where it goes downward it is decreasing.