# local max and local min values

• Mar 1st 2009, 11:02 AM
calcbeg
local max and local min values
Hi

The goal is to find the local max and local min values I get so far and then I don't know where to go - there are 2 questions - a and b

a) f(x) = x / (1+x)^2

f'(x) = (1-x)/(1+x)^3 according to my calculations which means that
x is underfined at -1 and f'(0) when x = 1 so

x= 1 is a critical point

f(1) = 1

but now what - is the local maximum and minimum both 1???

b) f(x) = x^3 - 3x^2 +5

f'(x) = 3x^2 - 6x
f'(x) = 3x(x-2)

so the critical points are x = 0 and x = 2

so f(0) = 5 and f(2) = 1 but how do you know which is local max and local min since I understand that sometimes the min is greater than the max

Help me understand - thanks

calculus beginner
• Mar 1st 2009, 11:11 AM
skeeter
for critical values where f(x) is defined, the following tests are used (also called the first and second derivative tests for extrema) ...

first derivative test ...

f(x) has a max if f'(x) changes sign from (+) to (-) at that critical value.

f(x) has a min if f'(x) changes sign from (-) to (+) at that critical value.

second derivative test ...

if f''(x) > 0 at the critical value, then f(x) is concave down and a max exists at that critical value.

if f''(x) < 0, then f(x) is concave up and a min exists at that critical value.