# Derivative, critical points...

• Oct 7th 2008, 09:40 PM
msimms3570
Derivative, critical points...
I have a polynomial -x^4/ 4-x^3/3+3 x^2 -11

Which I think I have found the correct derivative to.....-x^3-x^2-6x...is this correct?

If so, how do I find the critical points for this derivative, and also how do I figure out where f is increasing and where f is decreasing?
• Oct 7th 2008, 09:48 PM
Chris L T521
Quote:

Originally Posted by msimms3570
I have a polynomial -x^4/ 4-x^3/3+3 x^2 -11

Which I think I have found the correct derivative to.....-x^3-x^2-6x...is this correct?

If so, how do I find the critical points for this derivative, and also how do I figure out where f is increasing and where f is decreasing?

I presume that the function is $f(x)=-\tfrac{1}{4}x^4-\tfrac{1}{3}x^3+3x^2-11$??

If that's the case, then $f'(x)=-x^3-x^2+6x$

You find the critical points by setting $f'(x)=0$

So you should solve $-x^3-x^2+6x=0\implies -x(x^2+x-6)=0$

When you get your [3] critical points, set up a table [which should contain 4 intervals]. Pick a point within each of the intervals you set up and plug them into the derivative. The derivative will either have a positive or negative value at that point. Positive tells you that the function is increasing, and negative tells you that the function is decreasing.

Can you take it from here?

--Chris
• Oct 7th 2008, 10:29 PM
msimms3570
critical points
How do I figure out what the critical points are though?? Do I just plug in numbers that satisfy the equation equal to 0?
• Oct 7th 2008, 10:32 PM
Chris L T521
Quote:

Originally Posted by msimms3570
How do I figure out what the critical points are though?? Do I just plug in numbers that satisfy the equation equal to 0?

Treat this as if you're finding zeros for a function.

$-x(x^2+x-6)=0\implies -x(x+3)(x-2)=0$

This tells us that $x=-3$, $x=0$, and $x=2$.

So the intervals used in the first derivative test are $(-\infty,-3)$, $(-3,0)$, $(0,2)$, and $(2,\infty)$.

Pick a value from each of these intervals and plug it into $f'(x)$. If its positive, it increases on that interval, and if its negative, it decreases on that interval.

Can you take it from here?

--Chris
• Oct 8th 2008, 03:27 PM
msimms3570
points of inflection
Ok now I need help finding the points of inflection. I already took the second derivative which I found to be -3x^2-2x+6. Do I just set the equation to zero? Or do I have to use the quadratic formula?
• Oct 8th 2008, 03:33 PM
Chris L T521
Quote:

Originally Posted by msimms3570
Ok now I need help finding the points of inflection. I already took the second derivative which I found to be -3x^2-2x+6. Do I just set the equation to zero? Or do I have to use the quadratic formula?

In order to use the quadratic formula, you have to set it equal to zero. And since we're trying to find where $f''(x)=0\implies -3x^2-2x+6=0$, you will need to use the quadratic formula anyway, for I don't think that this can be factored.

--Chris
• Oct 8th 2008, 03:36 PM
msimms3570
Would the values the quadratic formula be a=-3, b=-2, and c=6?
• Oct 8th 2008, 03:39 PM
11rdc11
Yes
• Oct 8th 2008, 04:00 PM
msimms3570
Ok so I solve the quadratic formula and I get x=-1.786 and x=1.1196, however neither of these points are near any point of inflection that I can see on my calculators graph, and I'm pretty sure the graph on my calculator is correct. What did I do wrong???
• Oct 8th 2008, 04:13 PM
11rdc11
Hmm you got the correct values for the quad equation. a point of inflection is where a graph changes concavity so you could use the second derivative test to check the concavity. Test the interval $(-\infty,-1.786)~(-1.786,1.1196)~(1.1196,\infty)$
• Oct 8th 2008, 04:42 PM
msimms3570
test
how do I use the second derivative to test those intervals?
• Oct 8th 2008, 07:40 PM
11rdc11
Quote:

Originally Posted by msimms3570
how do I use the second derivative to test those intervals?

Just like you did for the 1st derivative test. See in what intervals it is positive and negative

This is from wikipedia

The second derivative test may also be used to determine the concavity of a function as well as a function's points of inflection.
First, all points at which are found. In each of the intervals created, is then evaluated at a single point. For the intervals where the evaluated value of the function is concave down, and for all intervals between critical points where the evaluated value of the function is concave up. The points that separate intervals of opposing concavity are points of inflection.