By now, someone may have become accustomed to the sum we formed *sum*time this Fall:

(1)

We’ve summoned this sum again in order to study its *extremum* behaviour, which we found to be somewhat fun. As is the norm, we’ll use the platform of calculus to this end and then give a summary of our findings just after *example 10*. See you there!!!

## Partial fraction decomposition

So we want to obtain the first two derivatives of the rational function (1). By default, the quotient rule should be used, but the fault here is that it gets a bit difficult — especially for the second derivative — and so we’ve chosen a different route.

If the denominator of a rational function is a product of *linear* factors, it may be suitable in some situations to first split the rational function into its partial fractions before proceeding with differentiation, rather than using the quotient rule. Depending on what one wants, the additional work done may be worth it in the end.

Resolve into its partial fractions.

The given rational function is *improper*, so we first put it in a proper form using long division:

Notice that . Set

and obtain the constants by the *cover-up* method:

Finally:

Find the *first derivative* of with respect to .

We make use of the partial fraction decomposition from the previous example to compute the derivative (compare this strategy with *logarithmic differentiation*).

A breeze.

Find the *second derivative* of with respect to .

Also a breeze because of the underlying work done in the initial partial fraction decomposition. Hard work pays.

## Polynomial factoring digression

Factorize .

The presence of some common factors makes the factorization a breeze.

See that special quadratic ? Get excited any time you spot it!!!

Find the *critical numbers* for .

E*q*uate the first derivative to zero and solve for :

The last line came from the factorization in the preceding example. Equate each factor to zero:

We obtain the *critical numbers*: . All *stationary/turning points*. The vertical asymptotes at are also critical numbers because the derivative is undefined at each of these, but these values are already ex*q*luded.

## Pretty familiar deductions

Show that give *local minimum values* for .

We use the *second derivative test*.

Since the second derivative is positive, gives a *local minimum*.

Since the second derivative is positive, gives a *local minimum*.

Show that give *local maximum values* for .

Starting with , let’s evaluate the second derivative.

Since the second derivative is *negative*, we conclude that gives a *local maximum* value for . Next, we calculate the second derivative when :

This may require a calculator to ascertain that it’s negative. Indeed, . Thus, gives a *local maximum* value for .

Find the *local minimum* value(s) and the *local maximum* value(s) for the rational function .

From the two preceding examples, give local minimum value(s) while give local maximum value(s). So we compute the corresponding values of at these points.

For , it helps to know that they are both solutions of . This makes the computation of easier, and we did that last post:

Thus, the *local minimum* value is and the *local maximum* value is . (Don’t be surprised that the local minimum value exceeds the local maximum value. They’re *local*, not *global*.)

Solve the equation .

We suspect that there is no real-valued solution to the given equation. Why? Because the right side exceeds the local maximum value of (and is smaller than the local minimum value of ).

Let’s confirm our suspicion. Clear fractions to get better traction and continue the solution:

The discriminant of the resulting quadratic is . Thus, there is no real-valued solution to the quadratic, and hence no real-valued solution to the given rational equation.

Cal*q*ulate for with vertices at , , .

The ingredients we need are the median-slopes:

and the side-slopes:

By definition and substitution:

Notice that the given triangle isn’t equilateral, yet satisfies . In our theory, once a triangle satisfies , then it shares some lateral properties with the equilateral triangle.

(In case you’re wondering how we obtained those coordinates, the building blocks are here in our second post of the year.)

## Takeaway

The following statements are *equivalent* for a triangle with side-slopes for sides :

- gives a local maximum value of
- or , where is the common ratio of the median-slopes
- the common ratio is that of an equilateral triangle (or its “look-alike”).

Okay, *q*ool.

## Tasks

- Show that the equation has no real-valued solution.
- Solve the equation .
- Given the rational function :
- Find the discriminant of the quartic numerator .
- Deduce that the graph of the rational function doesn’t cross the -axis.

- Find the interval(s) on which the rational function is
*increasing*, and the interval(s) on which it is*decreasing*. - Consider the rational function . PROVE that:
- its partial fraction decomposition is .
- the point is a
*local minimum*point. - the point is an
*inflexion*point.

- Consider the rational function . PROVE that:
- its partial fraction decomposition is .
- the point is a
*local minimum*point. - the point is an
*inflexion*point.

- PROVE that the following equations are equivalent:

(These equations play important roles in connection with triangles whose side-slopes form geometric progressions.)

- (Palindromic quartic) Verify that the roots of the quartic are:
- (Palindromic quartic) Find a monic quartic polynomial whose roots are:
- Show that if and only if

Notice that and are “multiples” of the *golden ratio*. Also, it may be more convenient to *solve* the given quartic, than to *substitute* the purported roots.

Our sum somehow leads to *palindromic polynomials*. We may visit this concept in the near *post*ure.