# Finding extrema when derivative has no rational roots.

• peripatein
In summary, extrema, also known as extreme values, are the highest or lowest points on a graph of a function, and they can be found in calculus by taking the derivative of the function and setting it equal to zero. If the derivative has no rational roots, the first or second derivative test can be used to determine the extrema. This involves finding the concavity of the function at a critical point. If the second derivative is positive, the function is concave up and the critical point is a local minimum. If the second derivative is negative, the function is concave down and the critical point is a local maximum. If the second derivative is zero, the test is inconclusive and another method must be used. It is possible
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)?

## 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.

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.

peripatein said:

## Homework Statement

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

## 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.

Last edited:
peripatein said:

## Homework Statement

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

## 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:

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

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

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

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

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

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

## 1. What are extrema in calculus?

Extrema, also known as extreme values, are the highest or lowest points on a graph of a function. In calculus, these points are found by taking the derivative of the function and setting it equal to zero.

## 2. How do you find extrema when the derivative has no rational roots?

If the derivative of a function has no rational roots, it means there are no points where the derivative is equal to zero. In this case, you can use the first or second derivative test to determine the extrema. The first derivative test involves plugging in values on either side of a critical point to see if the function is increasing or decreasing. The second derivative test involves finding the concavity of the function at a critical point.

## 3. What does it mean if the derivative has no rational roots?

If the derivative of a function has no rational roots, it means that there are no points where the slope of the function is zero. This could indicate that the function is either always increasing or always decreasing, or that it has a constant slope.

## 4. Can a function have extrema without having rational roots in its derivative?

Yes, a function can have extrema without having rational roots in its derivative. This occurs when the function has a constant slope or when the derivative is undefined at the critical points.

## 5. How do you use the second derivative test to find extrema?

The second derivative test involves finding the concavity of the function at a critical point. If the second derivative is positive, the function is concave up and the critical point is a local minimum. If the second derivative is negative, the function is concave down and the critical point is a local maximum. If the second derivative is zero, the test is inconclusive and another method must be used to determine the extrema.

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