Find your second derivative, then find for what values of x your second derivative equals 0. Calculate the f(x) with the values of x you find, the highest f(x) is your absolute maximum and the lowest is your absolute minimum.
find the absolute maximum and absolute minimum (if any) of the given function on the specified interval:
f(x)=x^2 + 4x + 5; -3=<x=<1
the book has:
max (1, 10)
min (-2, 1)
The book is a terrible resource, its not as clear as someone explaining it to you...as always I appreciate any help...thanks!
This one's easy cause u konw that this is a parabola that opens upward having only one minimum, You don't need calc, but if u insissssst.........
f'(x)=0 will give us all relative max/min.
but, it dont tell us nuttin bout if its a min or a max.....but the second derivative duzzzzzzzzzzzzz..................
So let's do it
- x = a is a locally a minimum if f"(a) > 0,
- x = a is a locally a maximum if f"(a) < 0; and
looks like we got ourselves only one max/min, but which one is it. I don't knoooooooooowwwwwwwww!!!!!!!!
Wait a sec.... Ill use the second derivative.
and 2>0! Its a minimum. YYAAAAAAAYYYYYYYYY!
its in the interval [-3,1] too! yyyaaaaaaaaaaaaaaaayyyyyyyyy!!!!
PEACE OUT!
ps We only found the min. The max can be found by evaluating and and see whichuns bigger.