Question

Hey, I'm working on some Analysis homework and I came upon a question prompting me to find critical points and I can't remember how. Can someone help?

f(x)=x^3+2x^2-4x+1

Answer 1

The critical points are any turning points in the graph. Where do they happen? Well, they happen anywhere the slope potentially changes sign (A), or anywhere the slope of the slope potentially changes sign (B).
Mathematically, how do you work that out? The slope potentially changes sign when the slope is 0, or undefined. The slope of the slope changes sign when the slope of the slope is 0, or undefined.

Answer 2

I get that. I'm having more trouble with the algebraic side of it.

Answer 3

Alrighty then. Let's take \[f(x)=x^3+2x^2-4x+1\], then\[\frac{d}{dx}f(x)=3x^2+4x-4\]Which has zeroes at 2/3 and -2 (that's just factorisation there. You can use the quadratic formula either).\[\frac{d^2}{dx^2}f(x)=6x+4\]which has zeroes at -2/3.

So the critical points are -2, -2/3, and 2/3

Answer 4

The part I don't recognize is how to derive a function. I'm only in Analysis, so I keep finding myself reading these complicated explanations I haven't learned yet. How do you derive a function to simply find the f'(x)?

Answer 5

It's quite strange that you haven't covered differentiation before doing this kind of analysis. I'm not sure how to do it otherwise without just plotting it, and figuring it out.

Answer 6

We probably have done it, but we're doing review for our final and I couldn't recall how to do it haha. Thanks anyway!

Answer 7

for an algebraic function:
eg f(x) = ax^n
f'(x) = anx^(n-1)

eg

2x^2: f'(x) = 2*2x^(2-1) = 4x^1 = 4x

x^3: f'(x) = 3x^2
5x: f'(x) = 5x^0 = 5*1 = 5

Answer 8

There's a lot of different parts to calculus, I'd recommend watching some of the videos on Khan Academy, starting with http://www.khanacademy.org/video/calculus--derivatives-1--new-hd-version?playlist=Calculus

Answer 9

Does that have something to do with power rule? Because that rings a bell.