# Quartic with two stationary points of inflection

1. Oct 1, 2012

### hmvince

Hey everyone!
Recently got a question in maths which asks:
"Use integral calculus to find the equation of the quartic that has stationary points of inflection at (1, 23) and (3, 15) and a y-intercept of 24"
This means that the second derivative has the form (as inflection points are x-intercepts in the second derivative):
f''(x) = k(x-1)(x-3)
I integrate this and get an answer for f'(x), all fine and dandy. But then I say, since there are two stationary points, f'(1) = 0, and f'(3) = 0, and it all breaks down!

Is it even possible to have a quartic with TWO stationary points of inflection, or am I just screwing something up (haha)?

Cheers.

2. Oct 1, 2012

### Mentallic

This should be in the homework section.

And sorry, but what is a stationary point of inflection? There are inflection points (f '' = 0), there are stationary points (f ' = 0), but a stationary point of inflection?

Anyway, I'll assume you mean just an inflection point. Of course quartics can have two stationary points, because clearly if you take the second derivative of a general quartic $f(x)=ax^4+bx^3+cx^2+dx+e$ you'll end up with a quadratic, which can have anywhere from 0 to 2 roots based on its discriminant.

3. Oct 1, 2012

### hmvince

Thankyou, and sorry, I should have explained. a stationary point of inflection is a point of inflection at which the rate of change is zero. Like this: http://mathworld.wolfram.com/images/eps-gif/InflectionPoint_700.gif

I've been sitting here trying to sketch a quartic with two of these points (of course a quartic with two points of inflection is possible ) and the only thing I can come up with is one which looks like a quadratic, but really fat. Like this: http://www.math.brown.edu/UTRA/polynomials/quartic.gif
where the central max/min forms a line with the other two, but I still cannot get the math to work.
$f'(x) = \int k(x-1)(x-3) = k(\frac{x^{3}}{3} - 2x^{3} + 3x) + c$
Then, f'(3) = 0 which results in:
$k(9-54+9) + c = 0$
$∴ c = 54k$
and f'(1) = 0
results in c = (4/3)k

How can c = both things, i dont understand! If k = 0, then I have nothing to integrate to get f(x)

Have I done something, or they have the question wrong?

Last edited: Oct 1, 2012
4. Oct 1, 2012

### pwsnafu

Show your working. Why would it break down?

Edit: argh cross post.

Why aren't you starting with a "generic" quartic as Mentallic post above, and differentiate twice, and solve for a and b?

Last edited: Oct 1, 2012
5. Oct 1, 2012

### pwsnafu

Err, no. An inflection point means that the second derivative changes sign. It's not enough that the second derivative vanish there.

6. Oct 1, 2012

### hmvince

Either way I do it, if I work back from the general form, I have unknowns on both sides, the k in front of one lot, and the a, b, and c in front of the second derivative of the general form.

Solving for A, b, and c I get:
k = 12a
k = -1.5b
k = (2/3)c

7. Oct 1, 2012

What's k?

8. Oct 2, 2012

### hmvince

$f''(x) = k(x-1)(x-3) = kx^{2} - 4kx + 3k$
and
$f''(x) = 12ax^{2} + 6bx + 2c$ (differentiating from general formula)

$∴12a = k$
$∴6b = 4k$
$∴2c = 3k$

9. Oct 2, 2012

### pwsnafu

Okay so substitute back into $f(x) = ax^4 + bx^3 + cx^2+dx+e$ and go from there.

10. Oct 2, 2012

### Mentallic

Ahh so it is. It's such a warm feeling knowing that I've had that wrong all these years.