Expert: Alon Mandes - 4/18/2009

Question
I nbeed to know not only the answer to the following questions bu t how to work them. What is the answer to determining where is the location of each local extremum of the function.

f(x) = x3 + 5x2 + 3x + 4

(2).      
 Find the largest open interval where the function is changing as requested.

Increasing f(x) = x2 - 2x + 1

(3).  Question 3     
 Solve the problem.

From a thin piece of cardboard 40 in. by 40 in., square corners are cut out so that the sides can be folded up to make a box. What dimensions will yield a box of maximum volume? What is the maximum volume? Round to the nearest tenth, if necessary.

 
 Find the largest open interval where the function is changing as requested.

Decreasing f(x) =sqrt of 4-x?

%5 Using the derivative of f(x) given below, determine the intervals on which f(x) is increasing or decreasing.

f'(x) = x1/3(x - 3)  

Answer
(1) f(x)=x³+5x²+3x+4. To find extremum we derive & solve the
   equation f'(x)=0 :
   f'(x)=3x²+10x+3
   3x²+10x+3=0 -> x1=-3 & x2=-⅓ . To decide whether x1 & x2 is
   min or max , we derive 2nd time & substitute x1 & x2 in f''.
   if f''>0 max , else min .
(2) f(x)=x²-2x+1=(x-1)². The function is increasing in the
   interval 1<x<∞
(3) Let's say that we cut x inches from each corner. The new
   dimensions will be now :
   Base=(40-2x)*40 INCH²,
   Height=x INCH .
   The volume of the open box will be : V=Base*Height
   V(x)=[(40-2x)*40]*x=(1600-80x)x=1600x-80x². To find min/max
   we derive: V'(x)=1600-160x . V'(x)=0 -> x=10 .
   V(10)=(1600*10-80*100)*10=80000 INCH³ . This is the max volume.
(4) f(x)=√(4-x) .
   This function is always decreasing for all x.
(5) f'(x)=(x-3)x^⅓.
   x=0 is a saddle point, & x=3 is min point (chick it!), so :
   Interval of increasing : x>3 ,
   Interval of decreasing : x<3 .