# calc

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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

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]

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