Inflection Points: How to Find Them and Why They Matter

[phillyolly]

Homework Statement

OK, this is a total disaster.

y=(2x³+x²+x+3)/(x²+2x)

Homework Equations

To find inflection points, we need to take the second derivative and solve it for x.

The Attempt at a Solution

I tried to find first derivative, and it looked horrible. The first derivative is

(2x⁴+8x³+x²-6x-6)/(x²+2x)²
So I am not sure if I made a mistake in doing the first derivative.
Or maybe my approach is totally wrong?...

 

[nonequilibrium]

Hm, I haven't tried it yet, but maybe writing it as f(x) + q(x)/(x²+2x) might help? (with q(x) of degree 1 or 0)

 

[LCKurtz]

Homework Statement

OK, this is a total disaster.

y=(2x³+x²+x+3)/(x²+2x)

Homework Equations

To find inflection points, we need to take the second derivative and solve it for x.

The Attempt at a Solution

I tried to find first derivative, and it looked horrible. The first derivative is

(2x⁴+8x³+x²-6x-6)/(x²+2x)²
So I am not sure if I made a mistake in doing the first derivative.
Or maybe my approach is totally wrong?...

Your first derivative is correct. But I don't think you will find the inflection point (there is one) without using a computer. The second derivative is a mess which simplifies to a fraction with a cubic in x numerator. This cubic has one real root which does in fact give an inflection point. But you aren't likely to find it by hand unless you know how, want to, and have the time to work through the hand algorithm for cubics. It makes me suspect that the problem is mistyped if it came from a calculus exercise section, or you are expected to use something like Maple to help you.

 

[tmccullough]

You can absolutely solve this by hand. First, follow mr. vodka's advice and apply long division, sp that you rewrite your function as
$$ax + b + \frac{cx+d}{x^2+2x}$$
I'm struggling to remember the name of the next step (partial fraction decomposition?), write it as
$$ax+b + \frac{g}{x+2} + \frac{h}{x}.$$

This is easy to take derivatives of any order, also each of these derivatives is set to zero fairly easily (how do you *easily* solve tx^n + u(x+2)^n = 0?). Work out these details for yourself.

I'd tell you the answer, but that's against forum policy I think.

 

[LCKurtz]

You can absolutely solve this by hand. First, follow mr. vodka's advice and apply long division, sp that you rewrite your function as
$$ax + b + \frac{cx+d}{x^2+2x}$$
I'm struggling to remember the name of the next step (partial fraction decomposition?), write it as
$$ax+b + \frac{g}{x+2} + \frac{h}{x}.$$

This is easy to take derivatives of any order, also each of these derivatives is set to zero fairly easily (how do you *easily* solve tx^n + u(x+2)^n = 0?). Work out these details for yourself.

I'd tell you the answer, but that's against forum policy I think.

You're right tmccollough. I didn't follow through vodka's hint and notice the special form you get.

 

[phillyolly]

Based on these suggestions, I tried to tweak the function. What I did was:
x(2x^2+x)/x(x+2) + (x+3)/(x^2+2x)=(2x^2+x)/(x+2)+(x+3)/(x^2+2x)

Is that what you are talking about?

Then, I solved the first derivative, then tried the second...Unsuccessful.

 

[Bohrok]

Do long division first so the degree of the numerator is less than the degree of the denominator. Then use partial fractions so that once you differentiate each term, you'll end up with something simpler than differentiating exactly what you started with.