# Math Help - Equation of a Function

1. ## Equation of a Function

I can't seem to figure out how to get the first part of this question done.

f(x)=x^3-x^2+4x-3
determine the intervals of increase or decrease

so I turned f(x) into derivative f'(x)=3x^2-2x+4 to try to find the critical points. using the quadratic formula to solve f'(x)=0 led me to a negative square root which I can't remember a way around and trying trial and error has lead me to further confusion. Or if anyone can tell me what to do with the square root of -44.

I have a feeling this is a simple answer and I'm just looking way too far into it, any help is appreciated!

2. Originally Posted by Tommelion
I can't seem to figure out how to get the first part of this question done.

f(x)=x^3-x^2+4x-3
determine the intervals of increase or decrease

so I turned f(x) into derivative f'(x)=3x^2-2x+4 to try to find the critical points. using the quadratic formula to solve f'(x)=0 led me to a negative square root which I can't remember a way around and trying trial and error has lead me to further confusion. Or if anyone can tell me what to do with the square root of -44.

I have a feeling this is a simple answer and I'm just looking way too far into it, any help is appreciated!
if $b^2 - 4ac < 0$, then $f'(x)$ has no real roots ... therefore no critical values.

$f'(x) > 0$ for all $x$, so $f(x)$ is increasing for all $x$.

3. Originally Posted by Tommelion
I can't seem to figure out how to get the first part of this question done.

f(x)=x^3-x^2+4x-3
determine the intervals of increase or decrease

so I turned f(x) into derivative f'(x)=3x^2-2x+4 to try to find the critical points. using the quadratic formula to solve f'(x)=0 led me to a negative square root which I can't remember a way around and trying trial and error has lead me to further confusion. Or if anyone can tell me what to do with the square root of -44.

I have a feeling this is a simple answer and I'm just looking way too far into it, any help is appreciated!
The reason that you aren't getting any real critical points from the derivative is because it never touched the x axis.

Therefore f(x) is always increasing