# Finding Intervals of increase and decrease

• Apr 6th 2010, 09:31 AM
Butum
Finding Intervals of increase and decrease
The Function:

f(x)=x^3-x^2+4x-3

Find the first and second derivative:

f'(x)=3x^2-2x+4
f''(x)=6x-2

I'm stuck on finding f'(x)=0, I have tried many ways, none have worked.

I have tried using the quadratic formula which does not work, because -2^2-(4(3)(4)) is a negative and I cannot square root it.

I don't know how to delete posts, skeeter solved this question about a month ago.

http://www.mathhelpforum.com/math-he...-function.html
• Apr 6th 2010, 10:57 AM
tonio
Quote:

Originally Posted by Butum
The Function:

f(x)=x^3-x^2+4x-3

Find the first and second derivative:

f'(x)=3x^2-2x+4
f''(x)=6x-2

I'm stuck on finding f'(x)=0, I have tried many ways, none have worked.

I have tried using the quadratic formula which does not work, because -2^2-(4(3)(4)) is a negative and I cannot square root it.

Of course the quadratic formula works and works marvelously! That the quadratic's discriminant $b^2-4ac$ is negative means there are no real roots and thus the function's derivative doesn't vanish at any real number and thus there are no extreme points.

Tonio

I don't know how to delete posts, skeeter solved this question about a month ago.

http://www.mathhelpforum.com/math-he...-function.html

.
• Apr 6th 2010, 11:20 AM
Butum
I never considered looking at it that way, thanks :)