How do I find the maximum of a function with a non-zero derivative?

Question

High school student here. I'm trying to find the maximum of this function: $$f(x)=\frac{2x-1}{2-x}.$$ where $0 \leq x \leq 1$. The standard process would involve finding the values of $x$ such that $f'(x)=0$ (then checking that the second derivative is negative), but $f'(x)$ here is always non-zero since $f'(x)=\frac{3}{(x-2)^2}$. From guessing and checking, I think that the maximum is probably when $x=1$, but this isn't particularly rigorous. How do I find the maximum of f(x) in this situation?

Answer 1

As you have found, $$f'(x)=\frac{3}{(x-2)^2}$$

This is always positive, but not defined at $x=2$.

You should check at the points of non-differentiability too, as this could mean a vertical asymptote. Thus a maxima could occur here.

If you check at $x=2$,

$$\lim_{x \to 2^-}=+\infty$$

This is clearly a point of maxima.

Moreover you could also have approached the problem as:

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

$$(y+2)(x-2)=-3$$

This is clearly a rectangular hyperbola (shifted) of the form $xy=k$..(k is some constant).

Its asymptotes are well known to be $$X=0 \ and \ Y=0$$

$$x-2=0 \ and \ y+2=0$$

Out of these x=2 is clearly a vertical asymptote.

You could also plot the function to find the maxima.

Take a look at the graph yourself here (Desmos)

Answer 2

If $f'(x)\gt0$ for all $x$ in an interval $[a,b]$, then $f$ is strictly increasing on that interval, hence its maximum (on that interval) occurs at the right endpoint, $x=b$. So there are two things to observe: first, that $3/(2-x)^2\gt0$ for all $x\not=2$, and second, that $2\not\in[0,1]$.