Finding extrema when derivative has no rational roots.

[peripatein]

Homework Statement

How may I find stationary points, increase/decrease intervals, concavity for f(x)=(x^3-2x^2+x-2)/(x^2-1)?

Homework Equations

The Attempt at a Solution

I am familiar with how it should be done, except that here I get f'(x)=x^4-4x^2+8x-1 for the numerator of the derivative and am unable to figure out how to find extrema. I'd appreciate some advice.

 

[haruspex]

Are you expected to find the exact locations of the stationary points, or merely whereabouts they lie? E.g. you can easily show there's one between 0 and 1.

 

[Ray Vickson]

Homework Statement

How may I find stationary points, increase/decrease intervals, concavity for f(x)=(x^3-2x^2+x-2)/(x^2-1)?

Homework Equations

The Attempt at a Solution

I am familiar with how it should be done, except that here I get f'(x)=x^4-4x^2+8x-1 for the numerator of the derivative and am unable to figure out how to find extrema. I'd appreciate some advice.

Can you see why there exists (at least) one positive root, and one negative root? Of course, there are two other roots, which are either both real or are complex conjugates of one another.

There are formulas for solving 4th degree polynomials, but the results are so complicated as to be almost useless. For example, Maple gives a positive root as

-1/6*6^(1/2)*((4*(82+9*83^(1/2))^(1/3)+(82+9*83^(1/2))^(2/3)+1)/(82+9*83^(1/2))^(1/3))^(1/2)+1/6*((48*(82+9*83^(1/2))^(1/3)*((4*(82+9*83^(1/2))^(1/3)+(82+9*83^(1/2))^(2/3)+1)/(82+9*83^(1/2))^(1/3))^(1/2)-6*((4*(82+9*83^(1/2))^(1/3)+(82+9*83^(1/2))^(2/3)+1)/(82+9*83^(1/2))^(1/3))^(1/2)*(82+9*83^(1/2))^(2/3)-6*((4*(82+9*83^(1/2))^(1/3)+(82+9*83^(1/2))^(2/3)+1)/(82+9*83^(1/2))^(1/3))^(1/2)+72*6^(1/2)*(82+9*83^(1/2))^(1/3))/(82+9*83^(1/2))^(1/3)/((4*(82+9*83^(1/2))^(1/3)+(82+9*83^(1/2))^(2/3)+1)/(82+9*83^(1/2))^(1/3))^(1/2))^(1/2)

and a negative root as

-1/6*6^(1/2)*((4*(82+9*83^(1/2))^(1/3)+(82+9*83^(1/2))^(2/3)+1)/(82+9*83^(1/2))^(1/3))^(1/2)-1/6*((48*(82+9*83^(1/2))^(1/3)*((4*(82+9*83^(1/2))^(1/3)+(82+9*83^(1/2))^(2/3)+1)/(82+9*83^(1/2))^(1/3))^(1/2)-6*((4*(82+9*83^(1/2))^(1/3)+(82+9*83^(1/2))^(2/3)+1)/(82+9*83^(1/2))^(1/3))^(1/2)*(82+9*83^(1/2))^(2/3)-6*((4*(82+9*83^(1/2))^(1/3)+(82+9*83^(1/2))^(2/3)+1)/(82+9*83^(1/2))^(1/3))^(1/2)+72*6^(1/2)*(82+9*83^(1/2))^(1/3))/(82+9*83^(1/2))^(1/3)/((4*(82+9*83^(1/2))^(1/3)+(82+9*83^(1/2))^(2/3)+1)/(82+9*83^(1/2))^(1/3))^(1/2))^(1/2)

Usually in such problems we just use numerical methods; that is why such methods were invented.

 

[SammyS]

Homework Statement

How may I find stationary points, increase/decrease intervals, concavity for f(x)=(x^3-2x^2+x-2)/(x^2-1)?

Homework Equations

The Attempt at a Solution

I am familiar with how it should be done, except that here I get f'(x)=x^4-4x^2+8x-1 for the numerator of the derivative and am unable to figure out how to find extrema. I'd appreciate some advice.

For the purposes of taking the 1st & 2nd derivatives, it may be useful to rewrite you function as:

\displaystyle f(x)=\frac{x^3-2x^2+x-2}{x^2-1}

\displaystyle =<br /> x-2+\frac{2(x-2)}{x^2-1}

\displaystyle =x-2+\frac{3}{x+1}-\frac{1}{x-1}​

 

[micromass]

For the purposes of taking the 1st & 2nd derivatives, it may be useful to rewrite you function as:

\displaystyle f(x)=\frac{x^3-2x^2+x-2}{x^2-1}

\displaystyle =<br /> x-2+\frac{2(x-2)}{x^2-1}

\displaystyle =x-2+\frac{3}{x+1}-\frac{1}{x-1}​

Very nice!

To the OP: this is a technique that always comes in very handy. If you want to see how SammyS did this, then you should research partial fraction decomposition. See http://en.wikipedia.org/wiki/Partial_fraction_decomposition#Examples