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 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