# critical numbers

• Nov 18th 2009, 02:14 PM
break
critical numbers
Find the critical numbers of the function.\$\displaystyle f(x) = x^3 + x^2 - x\$

whats the smaller value and the larger value?

the derivative of this is
3x^2+2x-1 but i get lost right after.
• Nov 18th 2009, 02:22 PM
skeeter
Quote:

Originally Posted by break
Find the critical numbers of the function.\$\displaystyle f(x) = x3 + x2 - x\$

whats the smaller value and the larger value?

the derivative of this is
3x^2+2x-1 but i get lost right after.

critical values occur where f'(x) = 0 or where f'(x) is undefined.

set the above derivative equal to 0 and solve for the critical values of x, then determine extrema.
• Nov 18th 2009, 02:30 PM
break
okay so

\$\displaystyle 3x^2+2x-1=0\$
\$\displaystyle 2x-1=3x^2\$
\$\displaystyle -1=3x^2/2x\$
\$\displaystyle -1=3x/2\$
\$\displaystyle 0 = (3x+2)/2\$

then

\$\displaystyle 0 = 3x+2\$
\$\displaystyle -2/3 = x\$

not sure if i got this part right.
• Nov 18th 2009, 02:41 PM
propjohn
Quote:

Originally Posted by break
Find the critical numbers of the function.\$\displaystyle f(x) = x^3 + x^2 - x\$

whats the smaller value and the larger value?

the derivative of this is
3x^2+2x-1 but i get lost right after.

f(x)=(x^3)+(x^2)-x
f'(x)=(3x^2)+(2x)-1

set that equal to zero. you will have to factor the polynomial to get

(3x-1)(x+1)=0

so

3x-1=0
3x=1
x=1/3

x+1=0
x=-1

those are your critical values
• Nov 18th 2009, 02:42 PM
propjohn
Quote:

Originally Posted by skeeter
critical values occur where f'(x) = 0 or where f'(x) is undefined.

set the above derivative equal to 0 and solve for the critical values of x, then determine extrema.

from my understanding, critical values are where f'(x)=0

partition numbers are where f'(x)=0 or is undefined
• Nov 18th 2009, 02:53 PM
skeeter
Quote:

Originally Posted by propjohn
from my understanding, critical values are where f'(x)=0

partition numbers are where f'(x)=0 or is undefined

Pauls Online Notes : Calculus I - Critical Points