Calculate the number of extremities for $f(x)=\frac{(|x|-1)\cdot (|x|-2)}{(|x|-3)(|x|-4)}$.

Question

I understand that graphing $g(x) = \frac{(x-1)\cdot (x-2)}{(x-3)(x-4)}$, for $x>0$ is enough to solve the question. So I require assistance in graphing g(x).
I have tried graphing f(x), but it was incomplete, I could get the following information:

1. roots of $g(x)$ are $2,3$

2. $\lim_{x\to 3^+} g(x) = - \infty$,
   $\lim_{x\to 3^-} g(x) = +\infty$,
   $\lim_{x\to 4^+} g(x) = +\infty$,
   $\lim_{x\to 4^-} g(x) = - \infty$,
   $\lim_{x\to \infty} g(x) = 1$

Without calculating the derivative and equating it to zero, is it possible to guess/calculate the points of extremas? Since even if I guess that there must be just 1 extrema between $x=1$ and $x=2$, I still wasn't able to get the information of this "hidden" extrema ($x=3.366$):

I was avoiding to try to calculate the derivative and equating it to zero, because it seemed tiresome, and I would like the quickest solution available to this.

Answer 1

Firstly note that $f(x)=g(|x|)$. If you can determine the stationary points of $g(x)$ then it is straightforward to deduce those of $f(x)$ by applying the modulus transformation as follows:

Given the graph of $g(x)$, erase all the graph for $x<0$ and reflect the graph of $g(x)$ for $x\geq0$ in the $y$ axis to obtain the entire graph of $g(|x|)$.

To find the stationary points of $g(x)$ without using calculus, set $$y=g(x)=\frac{(x-1)(x-2)}{(x-3)(x-4)}$$

Now rearrange this to form a quadratic in $x$, and you get $$(y-1)x^2+x(3-7y)+(12y-2)=0$$

For real $x$, the range of values of $y$ is determined by the discriminant of this quadratic, which must be greater than or equal to zero.

This leads to the inequality $$y^2+14y+1\geq0$$ with critical values $$y=-7\pm4\sqrt{3}$$

You can get the corresponding $x$ values by applying $$x=-\frac{b}{2a}=-\frac{3-7y}{2(y-1)}$$

Hereby you can obtain the exact coordinates of all the stationary points of $f(x)$

PS note that $-7-4\sqrt{3}=-13.92820323…$

Answer 2

I think you'll need some variation of the derivative.

For a small $\delta$, find $x$ so that $[g(x+\delta/2)-g(x)][g(x)-g(x-\delta/2)]$

$=g(x+\delta/2)g(x)-g(x+\delta/2)g(x-\delta/2)-g(x)^2+g(x)g(x-\delta/2)<0$

That should help you find where the derivative changes sign, roughly equivalent to where it is $0$. You might be able to use some inequalities to hone in on $x$.

The smaller your $\delta$ the more accurate your $x$.