Finding minima of the function.

Question

I am told to find minima of $$y=\frac{x}{4-x} +\frac{4-x}{x}$$

When I applied product rule to differentiate the eq. and equated to zero, using product rule, I got cubic equation and found three solutions which included two of them imaginary and one was real($=2$).

Whereas, while differentiating using quotient rule, I got only one ans.($=2$). I know imaginary was to be rejected at the end. But why my solution lacked those imaginary answers?

What if I had to get all three answers by differentiating using quotient rule and equating it to zero. I mean what manipulations I had to do in the intermediate equation. And what manipulations I had to do in the intermediate equation while differentiating the function using product rule to get one and only one answer instead of three.

My question is more about how equations really work and how they provide us with the solutions.

Any type of help would be thanked!

Answer 1

$$f(x)=\frac{4-x}{x} +\frac{x}{4-x}$$ $$f'(x)=\frac{32(x-2)}{(4-x)^2x^2}$$ $$f'(x)=\frac{-1}{x}-\frac{4-x}{x^2}+\frac{1}{4-x}-\frac{x}{(4-x)^2} =\frac{-x-(4-x)}{x^2}+\frac{(4-x)-x}{(4-x)^2}$$ $f'(x)=0$ than $x=2$;

Also let $g(x)=\frac{4-x}{x};$ then $$f(x)=g(x)+g(4-x)$$, $$f'(x)=g'(x)-g'(4-x)=0$$and this implies $x=2$

Answer 2

Since $y= z+\dfrac{1}{z}$, where $z=\dfrac{x}{4-x}$. Find out the critical points for the function $f(z)= z+\dfrac{1}{z}$, which would give you $z=\pm 1$ , analysing the second derivative of $f(z)$ would give you the minima at $z=1$. Substituting this value in your original equation would give the minima for your function as $$1=\dfrac{x}{4-x}\implies x=2$$