# absolute max and min

• May 29th 2009, 12:18 PM
lisa1984wilson
absolute max and min
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!
• May 29th 2009, 12:21 PM
mathieumg
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.
• May 29th 2009, 12:36 PM
VonNemo19
(Angry)
Quote:

Originally Posted by lisa1984wilson
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..................

• x = a is a locally a minimum if f"(a) > 0,
• x = a is a locally a maximum if f"(a) < 0; and
So let's do it

$\displaystyle f'(x)=2x+4=0\Longleftrightarrow{x=-2}$

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.

$\displaystyle f''(x)=2$ and 2>0! Its a minimum. YYAAAAAAAYYYYYYYYY!(Clapping)

its in the interval [-3,1] too! yyyaaaaaaaaaaaaaaaayyyyyyyyy!!!!(Clapping)

PEACE OUT!(Cool)

ps We only found the min. The max can be found by evaluating $\displaystyle \lim_{x\to{-3^+}}f(x)$ and $\displaystyle \lim_{x\to{1^-}}f(x)$ and see whichuns bigger.