### An algebra trick for finding critical points

The algebra trick here goes back at least $350$ years. This is worth looking at if only as an additional review of algebra, but is actually of considerable value in a variety of hand computations as well.

The algebraic identity we use here starts with a product of factors
each of which may occur with a *fractional or negative exponent.*
For example, with $3$ such factors:
$$f(x)=(x-a)^k\;(x-b)^\ell\;(x-c)^m$$
The derivative can be computed by using the product rule twice:
\begin{align*}
f'(x) &=k(x-a)^{k-1}(x-b)^\ell(x-c)^m+
(x-a)^k \ell(x-b)^{\ell-1}(x-c)^m\\
&\quad+
(x-a)^k(x-b)^\ell m(x-c)^{m-1}
\end{align*}
Now all three summands here have a common factor of
$$(x-a)^{k-1}(x-b)^{\ell-1}(x-c)^{m-1}$$
which we can take out, using the distributive law in reverse: we have
\begin{align*}
f'(x)&=(x-a)^{k-1}(x-b)^{\ell-1}(x-c)^{m-1}\\
&\quad \times
[k(x-b)(x-c)+\ell(x-a)(x-c)+m(x-a)(x-b)]
\end{align*}
The minor miracle is that the big expression inside the
square brackets is a mere quadratic polynomial in $x$.

Then to determine *critical points* we have to figure out the
roots of the equation $f'(x)=0$: If $k-1>0$ then $x=a$ is a critical
point, if $k-1\le 0$ it isn't. If $\ell-1>0$ then $x=b$ is a critical
point, if $\ell-1\le 0$ it isn't. If $m-1>0$ then $x=c$ is a critical
point, if $m-1\le 0$ it isn't. And, last but not least, *the two
roots of the quadratic equation*
$$k(x-b)(x-c)+\ell(x-a)(x-c)+m(x-a)(x-b)=0$$ are critical points.

*There is also another issue here, about not wanting to
take square roots (and so on) of negative numbers. We would exclude
from the domain of the function any values of $x$ which would make us
try to take a square root of a negative number. But this might also
force us to give up some critical points!* Still, this is not the main
point here, so we will do examples which avoid this additional worry.

#### Example

A very simple *numerical* example: suppose we are to find the *critical points* of the function
$$f(x)=x^{5/2}(x-1)^{4/3}$$ Implicitly, we have to find the critical
points first. We compute the derivative by using the product rule, the
power function rule, and a tiny bit of chain rule:
$$f'(x)={5\over 2}x^{3/2}(x-1)^{4/3}+x^{5/2}{4\over 3}(x-1)^{1/3}.$$
And now *solve* this for $x$? It's not at all a polynomial, and it
is a little ugly.

But our algebra trick transforms this
issue into something as simple as *solving a linear equation*:
first figure out the largest power of $x$ that occurs in *all* the
terms: it is $x^{3/2}$, since $x^{5/2}$ occurs in the first term and
$x^{3/2}$ in the second. The largest power of $x-1$ that occurs in
*all* the terms is $(x-1)^{1/3}$, since $(x-1)^{4/3}$ occurs in
the first, and $(x-1)^{1/3}$ in the second. *Taking these common
factors out* (using the distributive law ‘backward’), we rearrange to
\begin{align*}
f'(x)&={5\over 2}x^{3/2}(x-1)^{4/3}+x^{5/2}{4\over 3}(x-1)^{1/3}\\
&=x^{3/2}(x-1)^{1/3}\,\left( {5\over 2}(x-1)+{4\over 3}x\right)\\
&=x^{3/2}(x-1)^{1/3}\left({5\over 2}x-{5\over 2}+{4\over 3}x\right)\\
&=x^{3/2}(x-1)^{1/3}\left({23\over 6}x-{5\over 2}\right).
\end{align*}

*Now* to see when this is $0$ is not so hard: first, since the
power of $x$ appearing in front is *positive*, $x=0$ make this
expression $0$. Second, since the power of $x+1$ appearing in front is
*positive*, if $x-1=0$ then the whole expression is $0$. Third,
and perhaps *unexpectedly*, from the simplified form of the
complicated factor, if ${23\over 6}x-{5\over 2}=0$ then the whole
expression is $0$, as well. So, altogether, the *critical points*
would appear to be
$$x=0,{15\over 23},1.$$
*Many people would overlook the critical point ${15\over 23}$,
which is visible only after the algebra we did.*

#### Exercises

- Find the critical points and intervals of increase and decrease of $f(x) = x^{10}(x-1)^{12}$.
- Find the critical points and intervals of increase and decrease of $f(x) = x^{10}(x-2)^{11}(x+2)^{3}$.
- Find the critical points and intervals of increase and decrease of $f(x)=x^{5/3}(x+1)^{6/5}$.
- Find the critical points and intervals of increase and decrease of $f(x)=x^{1/2}(x+1)^{4/3}(x-1)^{-11/3}$.

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##### Calculus Refresher

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