Question

Finding critical points, after you find the zeros do I plug the test numbers into the derivative or the original function?

Answer 1

you are better post the whole problem to make clear the question.

Answer 2

ok

Answer 3

but i think you put the critical point into the original one to figure out the value of f and then compare them together

Answer 4

don't forget the ending points. i mean interval

Answer 5

f(x)=-1/x^2 on the interval from .5<=x,=2 f '(x) =-x^2x^-3
then set to 0 ssolve for x, x=0. so I know I plug in the end points and x values to find max/min. But do I plug into the derivative or the original function?

Answer 6

the interval is not clear, \[0.5\le x \le 2\]is it right?

Answer 7

yes sorry

Answer 8

and your f'(x) must be = 2/x^3. tell me if i am wrong. I 'm dealing with critical point, too. we can discuss and help each other

Answer 9

ok great! I think its 2x^-3 though

Answer 10

or 2 1/x^3

Answer 11

why? your f(x) = -x^-2. take derivative of it, ok, you are right now, not previous result

Answer 12

so, you don't have critical point except ending point. because x cannot be 0 to make f'x defined

Answer 13

oh that makes sense ,

Answer 14

count f(0.5) and f(2) and compare them only

Answer 15

So then x wont have a 0 then because its in a denominator..... ok I see so then that means that i doesn't cross th x axis then right?

Answer 16

it doesn't mean that. the critical points talk about max, min of function

Answer 17

But then do you use f(.5) or f ' (.5) ?

Answer 18

f(0.5)= -2 is local min and f(2)=-1/4 is local max

Answer 19

oh right this tells me when the tan line is =0,

Answer 20

if consider of the value on interval [0.5,2] they are absolute max and min

Answer 21

not crossing the axis

Answer 22

i think you confuse. the critical point tells us about the tangent line is horizontal line with the slope =0. in your case, you don't have it. if you have it is vertical line with x =0

Answer 23

that is y-axis

Answer 24

in other word, it has y axis as its asymtote

Answer 25

verify by drawing the graph, trivially see that

Answer 26

Hello, do I lost you?

Answer 27

i'm back sorry

Answer 28

so i plot my intervals on line, then plug test numbers in to each interval right?

Answer 29

yes

Answer 30

ok so f'(-1)=-2 to the left of .5,dec.
f'(1)=2 in between .5 and 2 inc
and f'(3)=2/27 to rt of 2 = inc

Answer 31

why you put those points into f'?. just 2 ending points into ORIGINAL function

Answer 32

so i should put the end points into the original to test?

Answer 33

not to test, to figure out which point is maximum, or minimum value of f

Answer 34

? the first derivative test to see if inc/dec right

Answer 35

yes, of course. if you have f(0.5)> f(2). does it mean your function decreases?

Answer 36

so after the first derive test then you put the x's into original to find the corresponding y values?

Answer 37

yes

Answer 38

ok i see i'm gonna try another one and get back to you thanks

Answer 39

the derivative just to make sure that among the interval, your function has another up or down or not. if it has another critical points, put those points to ORIGINAL one to know it it's up or down

Answer 40

no critical point, f decreases from a, to b.