Finding intervals of a 3 degree function?

[KevinFr]

The question says find apex, low point and the monotonic properties of the functions. a) b) c)...

To find intervals, I use the abc-formula. Example:

f(x) = 3x^3 - 3x
d/dx * f(x) = 3 * 3x^2 - 3, here a=3*3, b= -3 and c=0 (because there is none)

x1 = ( -b + sqrt(b^2 + 4*ac) ) / 2a
x2 = ( -b - sqrt(b^2 + 4*ac) ) / 2a

So from this I get 3 intervals:

[-infinity, x1], [x1, x2], [x2, +infinity]. Out of these intervals I can find out where graph rises and falls, in a 2 degree function.

In a 3 degree function there should be 4 intervals, correct? What would be the best approach to find all of them? I'm sorry if it's a stupid question, but I'm really struggling with this...

In the .pdf file you can see how I find the intervals (written as intervaller) and if the graph rises or lowers (written as stigning). Avoid the text as it's written in Norwegian.

 

[BvU]

Hello Kevin,

For a 3 degree function you can find a maximum of 4 intervals: e.g. $f(x) =x^3$ doesn't have 4.
You appear to be solving ${df\over dx} = 0$ and find something you don't seem to check ?
Something is going wrong there...

I am referring to your example $f(x) = x^3 - 3x$

For homework exercises, we have an excellent template that would be very helpful here! It goes like:

Homework Statement

Homework Equations

The Attempt at a Solution

​

A further comment: I should think that you are supposed to be able to decompose the f(x) from the example and the f(x) from 7.70 in factors easily. Both can be written in the form of (x-a)(x-b)(x-c) c.q. (x-a)(x-b). If not, practice !

 

[BvU]

Looking at your pdf:
you have $f(x) = -x^2+4x+3$ and differentiate: $f'(x) = -2x+4+0$. $f'(x) = 0$ is not an equation to tackle with the abc formula. (Is that what one calls a golden hammer for you ? ) $\quad 4-2x = 0 \Rightarrow x = 2$ and that's it. To see if it's a minimum or a maximun (if you haven't seen that already from the $-x^2\$) : $\quad f''(2) = -2 < 0$ so a maximum.

The case $f(x) = 0$ can be tackled with the abc formula but it is much easier to look 'through' and write $f(x) = 0 \Leftrightarrow x^2-4x+3 = 0$. To write this as $$(x-a)(x-b) = x^2 - (a+b) x + ab$$ we are searching for two numbers a and b that add up to 4 and their product is 3. How hard can it be ?

Another way to write $ax^2 + bx + c$ as a product is indeed using the abc formula. Do you know where it comes from ?

 

[KevinFr]

The case f(x)=0f(x)=0f(x) = 0 can be tackled with the abc formula but it is much easier to look 'through' and write f(x)=0⇔x2−4x+3=0f(x)=0⇔x2−4x+3=0f(x) = 0 \Leftrightarrow x^2-4x+3 = 0 . To write this as

(x−a)(x−b)=x2−(a+b)x+ab​

Aaah. Thank you for the explanation! That cleared it up for me :D

 

[BvU]

Good. I take it you came to the right answers then.

PS:
Knowing how to derive the abc thingy is useful throughout a career ...