Ummm 48 cm ima be honest I need some points my fault
THE ANSWER TO THAT PROBLEM IS SIMPLY 175
The local minimum of function is an argument x for which the first derivative of function g(x) is equal to zero, so:
g'(x)=0
g'(x)=(x^4-5x^2+4)'=4x^3-10x=0
x(4x^2-10)=0
x=0 or 4x^2-10=0
4x^2-10=0 /4
x^2-10/4=0
x^2-5/2=0
[x-sqrt(5/2)][x+sqrt(5/2)]=0
Now we have to check wchich argument gives the minimum value from x=0, x=sqrt(5/2) and x=-sqrt(5/2).
g(0)=4
g(sqrt(5/2))=25/4-5*5/2+4=4-25/4=-9/4
g(-sqrt(5/2))=-9/4
The answer is sqrt(5/2) and -sqrt(5/2).
So the first hour it travels from 1-6 miles, the second hour it travels at least two miles. And the question is "draw the range of miles the boat could of traveled in the second hour". Well first of all the third hour is completely useless so forget about that. Now, the first hour says it traveled from 1-6 miles, we don't know how many miles it traveled for sure but it could of traveled up to 6. So in the second hour when they say "at least two miles" the boat could of travled at least two miles more than 1-6. So 1+2+3, 2+2+4, 3+2=5, 4+2=6, 5+2=7 and 6+2=8. So the boat could of traveled from 3 to 8 miles in the second hour.
