Posted by **kooldude** on Thursday, November 9, 2006 at 11:05am.

one more question!!

what values of c deos the polynomial f(x) = x^4 + cx^3 + x^2 have two inflection points? one inflection point?

At the inflection point, the slope of the curve is zero.

Find the dervative of f(x) and set it equal to zero. For certain values of c, there will be one or two values for x.

isn't it the second derivative

I apologize. My initial response was completely wrong.

You are correct. The

*second* derivative will equal zero at inflection points. Not the first.

okay so i got this

2nd der = 12x^2 + 6cx + 2 = 0

i get stuck here

Assuming your second derivative is correct, setup the quadratic formula.

You will find that for certain values of c, x can have one or two values.

okay, so trial and error?

At an inflection point the second derivative is zero. The second derivative is:

f''(x) = 12x^2 + 6cx + 2

If this quadratic function is to have only one zero, it must be of the form:

A (x-B)^2

The coefficient of x^2 is A but you know that this must be 12, so A = 12. The constant term is A B^2 = 12 B^2, which must be 2, so you can deduce that that B = +/- squareroot[1/6]. The term linear in x is

-2ABx = -/+ 24 squareroot[1/6] x

This must be 6cx, so you find that

c = -/+ 4 squareroot[1/6]

So, you see, this problem can again be solved using differentiation and factorization. These sort of problems are a simple and boring. Do you want to see a more interesting application of these techniques?

No. Setup your quadratic formula. The formula includes a plus-minus sign, which means there are zero, one or two solutions for each quadratic.

In other words, solve for x.

At an inflection point the second derivative is zero. The second derivative is:

f''(x) = 12x^2 + 6cx + 2

If this quadratic function is to have only one zero, it must be of the form:

A (x-B)^2

The coefficient of x^2 is A but you know that this must be 12, so A = 12. The constant term is A B^2 = 12 B^2, which must be 2, so you can deduce that that B = +/- squareroot[1/6]. The term linear in x is

-2ABx = -/+ 24 squareroot[1/6] x

This must be 6cx, so you find that

c = -/+ 4 squareroot[1/6]

what is BOZOOOOOOOOOOOOOOOOOOOOOOOOOO

## Answer this Question

## Related Questions

- Calculus - The number of people who donated to a certain organization between ...
- calculus - Find all relative extrema and points of inflection of the function: f...
- Calculus - If f(x) is a continuous function with f"(x)=-5x^2(2x-1)^2(3x+1)^3 , ...
- calculus - Use a graph of f(x) = 3 e^{-8 x^2} to estimate the x-values of any ...
- Math - exponential - How many inflection points does f(x) = xe^(x^(-2)) have...
- calculus - f(x)= x^3/x^2-16 defined on the interval [-19,16]. Enter points, such...
- Calc - Consider the cubic polynomial y=Ax^3 +6x^2 -Bx, where A and B are unknown...
- Math - Assume k is a constant. the function g(x)= x4- 6k(x2) + 6k2 + x has at ...
- Calculus please help me - f(x) = \frac{ x^3 }{ x^2 - 25 } defined on the ...
- Math(Please help) - 1) where is the function f(x)= 5 / (2x-4)^3 increasing. I ...

More Related Questions