Finding extremum of quadratic function divided by $x$

Question

How can we find the extremum value for the function: $$f(x) = \frac{-(3x-4)^2}{2x}$$

I plugged this into GeoGeBra and found two extremums, $24$ and $0$, but how can we find it without using a calculator?

I've thought about calculating the extremum of the numerator first, then dividing it by the respective $x$ value, but it obviously don't work in this case.

How can we solve it?

Answer 1

Here is a solution without calculus.

For $x>0$, $\dfrac{(3x-4)^2}{2x}\ge0$, so $f(x)\le0$,

with $f(x)=0$ when $3x-4=0$.

For $x<0$, $(3x+4)^2\ge0$, so $(3x-4)^2\ge-48x$, so $f(x)\ge24$,

with $f(x)=24$ when $3x+4=0$.

Answer 2

$f(x)=\displaystyle-\frac{(3x-4)^2}{2x}$

$\displaystyle\frac{d}{dx}f(x)=-\frac{d}{dx}\frac{(3x-4)^2}{2x}$

$\displaystyle\frac{d}{dx}f(x)=-\frac{2x\frac{d}{dx}(3x-4)^2-(3x-4)^2\frac{d}{dx}2x}{4x^2}$

$\displaystyle\frac{d}{dx}f(x)=\frac{2(3x-4)^2-12x(3x-4)}{4x^2}$

$\displaystyle\frac{d}{dx}f(x)=-\frac{(3x-4)(6x+8)}{4x^2}$

For maximum and minimum values of $f(x)$, we have

$\displaystyle\frac{d}{dx}f(x)=0$

$\displaystyle-\frac{(3x-4)(6x+8)}{4x^2}=0$

$x=\frac{4}{3}$ or $ -\frac{4}{3}$

Clearly $x=\frac{4}{3}$ gives$f(x)=0$ and $x=-\frac{4}{3}$ gives $f(x)=24$ which is as you found graphically.