1. ## calculus

Hello everyone. I hope all are well. This is my first post and visit of many to come in the future I'm sure. My name is Nav and I am currently a college student that went back to school to get my degree.

I was hoping someone could help me with a couple of homework problems I have. I did most of my homework, but I couldn't get these last two. I believe the teacher purposely made these last two hard.

1. Show that every (degree 3) polynomial has exactly one point where its second derivative is zero. Call this point P.

2. The show if a polynomial has three x-intercepts x1, x2, and x3, then x-coordinate of P is equal to x1+x2+x3/3.

Any help would be greatly appreciated. I am taking calculus after 10 years so I am having a bit of difficulty, but I am learning and picking it up again. I didn't know where else to go. Thank you for your time. Thank you, thank you, thank you, thank you..lol

2. Originally Posted by omniconsum3r
...

1. Show that every (degree 3) polynomial has exactly one point where its second derivative is zero. Call this point P.

2. The show if a polynomial has three x-intercepts x1, x2, and x3, then x-coordinate of P is equal to (x1+x2+x3)/3.

Any help would be greatly appreciated. I am taking calculus after 10 years so I am having a bit of difficulty, but I am learning and picking it up again. I didn't know where else to go. Thank you for your time. Thank you, thank you, thank you, thank you..lol
1. Let
$\displaystyle f(x)=ax^3+bx^2+cx+d$ be the function of third degree.

Then the second drivative is:

$\displaystyle f''(x) = 6ax+2b$ . With f''(x) = 0 you'll get: $\displaystyle x = -\frac b{3a}$

Calculate $\displaystyle f\left( -\frac b{3a} \right) = -\frac{bc}{3a} + \frac{2b^3}{27a^2} + d$ ....... Therefore $\displaystyle P\left( -\frac b{3a}~,~ -\frac{bc}{3a} + \frac{2b^3}{27a^2} + d\right)$

2. If the function has 3 zeros you can write f as a productof the linear factors:

$\displaystyle f(x)=a(x-x_1)(x-x_2)(x-x_3) = a(x^3-x_1x^2-x_2x^2-x_3x^2 + uninteresting\ summand)$ The uninteresting summand contains x to the power of 1 or 0.

Then the second derivative is:

$\displaystyle f''(x)=a(6x-2(x_1+x_2+x_3))$ With f''(x) = 0 you'll get the given result.

3. You are awesome dude!!! Thank you man, thank you. I understand what you did which is even more awesome!