Graphing Polynomials: f(x)= x^4 - 3x^3 - 3x^2 + 11x - 6

[npaggs]

I seem to be getting stuck on the second derivative.

Here's what I've gotten etc

Original equation:
f(x)= x^4 - 3x^3 - 3x^2 + 11x - 6

Used long division/synthetic division, whatever term you're accustomed to and received the answer x^3 - 2x^2 - 5x + 6 along with the root (x - 1)

I then used long division again with that equation and got the answer x^2 - x - 6 with corresponding roots (x - 3) and (x + 2)

Now that that's settled, I continued on to the first derivative.

I figured f^1(x) = 4x^3 - 9x^2 - 6x + 11

I then first started out with long division, getting the equation 4x^2 - 5x - 11, then I put that equation into the quadratic formula and found that the local extrimas = (2.4, -5.2) and (-1.2, -16.3)

now for the second derivative i got
f^2(x) = 12x^2 - 18x - 6

Now I have no idea where to go from here, it seems that nothing is working, have I made a sign error some where? I can't figure it out.

Any help would be much appreciated

 

[Deleted member 4993]

Re: Graphing Polynomial Equations

I seem to be getting stuck on the second derivative.

Here's what I've gotten etc

Original equation:
f(x)= x^4 - 3x^3 - 3x^2 + 11x - 6

Used long division/synthetic division, whatever term you're accustomed to and received the answer x^3 - 2x^2 - 5x + 6 along with the root (x - 1)

I then used long division again with that equation and got the answer x^2 - x - 6 with corresponding roots (x - 3) and (x + 2)

Now that that's settled, I continued on to the first derivative.

I figured f^1(x) = 4x^3 - 9x^2 - 6x + 11

I then first started out with long division, getting the equation 4x^2 - 5x - 11, then I put that equation into the quadratic formula and found that the local extrimas = (2.4, -5.2) and (-1.2, -16.3)

now for the second derivative i got
f^2(x) = 12x^2 - 18x - 6

Now I have no idea where to go from here, it seems that nothing is working, have I made a sign error some where? I can't figure it out.

Any help would be much appreciated

You have not told us - clearly - what did the problem ask for.

 

[Loren]

Re: Graphing Polynomial Equations

I think you are okay.
When you factored the original equation you got (x-1)[sup:3bp0n3n5]2[/sup:3bp0n3n5](x+2)(x-3). This gives you a double intercept at (1,0), and single intercepts at (-2,0) and (3,0).

Your first derivative gives you slope at any point on the curve. To find the relative max and mins set your first derivative to 0 and solve for x. I think you will get zeros at x=1 and \(\displaystyle x=\frac{5\pm \sqrt{201}}{8}\). This tells you that the curve is at a maximum or a minimum when x = those three values. plug each of those values into the original equation to get the exact max and min points.

Set your second derivative = to zero and solve for x. This gives you the values of x at which there is an inflection point.

The first derivative is useful in telling the slope of the curve at any given point. For instance at the x-intercept (-2,0), substituting -2 into the first derivative you get -45. This tells you that the curve is going down pretty darn steeply as it passes through (-2,0).

If you have a graphing calculator, it will help you see what's going on. However, don't come to rely on the calculator. Understand how these derivatives work and what they tell you, then check it out on the calculator.

Hope this helps