# Math Help - features of a polynomial function

1. ## features of a polynomial function

The the function f(x)=x^3-x^2+4x-3, determine
a) the intervals of increase or decrease
b) the location of any maximum or minimum points
c) the intervals of concavity up or down
d)the location of any points of inflection

I have gotten this far
the derivatives are
f'(x) = 3x^2 -2x +4
f''(x)= 6x-2

I cant figure out how to factor the first derivative in order to find the zeros, help anyone?

2. Originally Posted by surffan
The the function f(x)=x^3-x^2+4x-3, determine
a) the intervals of increase or decrease
b) the location of any maximum or minimum points
c) the intervals of concavity up or down
d)the location of any points of inflection

I have gotten this far
the derivatives are
f'(x) = 3x^2 -2x +4
f''(x)= 6x-2

I cant figure out how to factor the first derivative in order to find the zeros, help anyone?
note that the discriminant $b^2-4ac < 0$ ... that tells you that $f'(x)$ is never equal to 0 for real values of x.

so ... if $f'(x) \ne 0$ , that means $f'(x)$ can be one of two possibilities ... (you complete the deduction).

3. okay so I'm trying to solve for the intervals of increase and decrease for the graph, but does that mean it doesn't increase or decrease at any point if I cannot factor its derivative? I'm just following the steps in the book and it says to find the zeros in order to solve this

4. f'(x) has no real zeros ... therefore f'(x) is either always positive or always negative ... which further means that f(x) is always increasing or always decreasing.

how can you tell?

5. i graphed it on my graphic calculator and the function does have a slight curve to it, but I have no idea how to solve for this mathematically, I'm assuming there is some way to do it since the question has multiple parts but it just doesn't make any sense to me. All the examples in my text use polynomials that can be factored

6. $f'(x) > 0$ for all $x$ ... this says that $f(x)$ is increasing for all $x$ ... there are no maximums or minimums.

$f''(x) = 0$ at $x = \frac{1}{3}$. since $f''(x)$ changes sign at $x = \frac{1}{3}$ , $f(x)$ has an inflection point there.

for $x < \frac{1}{3}$ , $f''(x) < 0$ ... $f(x)$ is concave down on this interval.

for $x > \frac{1}{3}$ , $f''(x) > 0$ ... $f(x)$ is concave up on this interval.

look at the attached graph and confirm the analysis ...