# Math Insight

### 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

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