Calc whizzes

[CoCoNuT]

f<x> = x^4 + 5x^3 + 2

just looking quickly it should have 3 sign changes on the derivitive(4x^3 + 15x^2). To find the turn around points you factor f'(x) you get x^2(4x+15) and set it equal to 0. since it mulitplication and equals 0: x^2 = 0 ; 4x + 15 = 0. So we find 2 turn around points : 0 and 3.75 respectivly. Where did the 3rd one go? My thinking is there is a positive and negative 0?

[Chris]

not properly looked at it, but one or more of the roots you've found may be real roots, i.e. they lie on the x-axis itself, and so can be counted twice. plot the graph and see what it looks like, may help a lot.

[Mark]

4x + 15 = 0
4x = - 15
x = -3.75

Actually I have no idea what you mean by 3rd one.. could be my understanding of Maths in English

[simu]

there are three xs...

look at it this way:

f'(x) = x(x(4x+15))

so you get:

1. x = 0
2. x = 0
3. x = -3.75

the first two xs are the same, which means that there is a minima or maxima.

[aspfreakout]

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