Thread: 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

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 $\displaystyle b^2-4ac$ 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

.

I never considered looking at it that way, thanks