Question

hey can some one give me the steps in order on locating critical points

Answer 1

You look to see where the derivative changes directions (from postive to negative or vice versa) or where the limit doesn't exist.

Answer 2

find derivative, and look wher it becomes 0

Answer 3

ok i found the deriv , do i enter zero in for x next ?

Answer 4

no. Equal to 0 and solve to find the x

Answer 5

later check the sign to the left and right. If it changes you got your point. If not, just an inflection.

Answer 6

ok this is what i have 4x^3-4=0 for deiv

Answer 7

deriv

Answer 8

this is derivative already?

Answer 9

yes i put it in the deriv from the orign equat x^4-4x

Answer 10

x^3-1 =0
so x = 1

Answer 11

ok how did you get there

Answer 12

now: for x<1 derivative is negative
for x>1 positive
so you got minimum

Answer 13

4x^3-4 devide by 4 and make equal to 0

Answer 14

ok its saying consider the functions f(x)=x^4-4x
step 1 find critical points
step 2 find open interval of which function f(x) is incr and decres
step 3 find local relative extrema
step 4 find absolute extrema

Answer 15

step 4 interval is closed

Answer 16

x<1 function decreasin
x=0 local minimum and also abolut minimum
x>1 function increasing
for absolute extrema you will have the values that the function gets on the limiting points of your interval [a,b]. F(b), f(b) - absolut maximums

Answer 17

Sry, made typing mistake:
f(a), f(b) - absolut maximums

Answer 18

ok its fine , is it possible to graph this equation and i can solve from graph

Answer 19

http://www.wolframalpha.com/input/?i=x%5E4-4x

Answer 20

thank you so much