find the critical point to determine the value of max and min
from :
f(x)=x^3-3x+1 on the interval (-(3/2), 3)

Question

find the critical point to determine the value of max and min
from : 
f(x)=x^3-3x+1  on the interval  (-(3/2), 3)

anonymous · Answer

first find the derivative of this function

anonymous · Answer

then set the serivative to zero, and solve, these will be posibble critical points

anonymous · Answer

find the derivative and set it to zero
3x^2-3=0
solve it we have 3x^2=3
x=1 or z=-1
They are your max. and min. values.

anonymous · Answer

so how about the interval (x,y)=(-(3/2), 3)

anonymous · Answer

but, be careful, b/c since you were working with an interval, you must evluate the endpoints , in this case-3/2 and 3. If you find that these are greater or smaller, that the value you found setting the derivative to zero, then these would be your max or min

anonymous · Answer

may be u can describe detail..?

anonymous · Answer

right, so first the other guy got you the critical values, by setting the derivative to zero.
You got 1 and -1. What you have to do is evluate these points, plug them into the orginal function. Set those points aside. Now evluate the endpoints of the interval you are working with(plug them into the orginal function). Now compare these values witht he values you got from setting the derivative to zero.

anonymous · Answer

The highest max and the lowest min are the absolute max and min. The rest are just local max and min