sensation612

there has to be someone that have taken calc and passed, so...if u know please help me out b/c i'm screwing up in this class

"Show all of your work including critical numbers, like critical numbers and the number lines used to determine the answers."
(dotted line means "over" and ^ mean powered)
f(x)= 2x^2
---------
x^2-4

f '(x)= -16x
--------------
(x^2-4)^2

f "(x)= 16(3x^2 + 4)
-----------------
(x^2 - 4)^3

A) find all relative extrema.

B)Find all points of inflection.

Nefir

Ok I am not going to solve this problem for you, but I'll do my best to point you in the right direction so you can solve it yourself. Forgive me if I miss something or completely butcher it... my calc skills are way past their "sell by" date

Lets start with the obvious, and you can put 2 and 2 together

a)
-Relative extrema just means relative maximum or minimum. One of these two extrema will occur when the rate of change goes from positive to negative or vice-versa.
-In order for that to happen, the rate of change must hit zero.
-The first derivative is the rate of change. Solve for 0.
-Once you find such points, you need to make sure whether each is a maximum or a minimum.

b)
- Points of Inflection are points where the curve changes from curving up to curving down, or vice-versa.
- This means somewhere between curving up and curving down, the function is NOT curving anywhere, ie: the rate of change is 0.
- Rate of change is the first derivative. Solve for 0.

Good luck

PS: Critical point = min, max, or inflection point.

PPS: I DID butcher it! Fixed now.. I hope.

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phukraut

how to find a min/max of f(x):

take the first derivative of f(x), call it f'(x). find all the points where f'=0. plot them on a number line for x. pick a number between these points, and plug that number into f'(x) and see if the result is greater than or less than zero. do that for all the intervals between your "zero"-points. this will tell you where the curve goes up and down.

example: f(x)=x^2. f'(x)=2x. f'(0)=0, so this is a critical point. try f'(-10)=-20<0. f'(10)=20>0.


f'<0 here ||| f'>0 here
---------0-----------.

if a curve goes up (f' greater than zero), then reaches a zero point (f'=0), then goes down (f' less than zero), you have a local maximum. if it goes down, then zero, then up, then you have a local minimum.

to find out what the actual max/min point is, you plug the zero-point into f(x) (the original function), and get a y coordinate. for example, if f'(M)=0 is your only zero point, and f'(M-1)>0 and f'(M+1)<0, then you have a local maximum at the point (M, f(M)).

this process is caleld the First Derivative Test.

points of inflection have to do with how the curve goes up or down. is it concave up or concave down? for example, does it go

\___/, (concave up)

or does it go
___
/ \? (concave down)

in order to find out, you take the second derivative, and find out where it is zero. if f''>0, then f is concave up. if f''<0, then f is concave down. you can also find min/max via concavity. if f'=0 at an extreme point, and f''>0, then f has a local min at this point. if f'=0 and f''<0, then f has a local max. this is called the Second Derivative Test. an inflection point is where the curve goes from concave up to concave down or vice versa. to find the point of inflection, determine where the concavity changes in x, and plug that value into f(x) to get the y coordinate.