# To show $f(x)$ has ONLY one Max in $x\in[0,1]$

I have function $$f(x)=\left(\frac{1-x}{2-x}\right)x^{p-1}~\text{where}~p>1;~x\in[0,1]$$ I want to show that $f(x)$ has ONLY one maximum in $x\in[0,1]$.

I get the second derivative as $$f"(x)=\frac{x^{p-3} \left[p^2 (x-1) (x-2)^2-p (x (3 x-7)+6) (x-2)+2 (x-1) ((x-2) x+4)\right]}{(x-2)^3}$$ But it is difficult to show $f"(x)<0;~x\in[0,1]$, because I could not show that the term in squared brackets is positive (as denominator negative). Anyone can show it???

However, I noticed that $\left(\frac{1-x}{2-x}\right)$ and $x^{p-1}$ are monotonically decreasing and increasing functions, respectively, in $x\in[0,1]$. Can these properties be helped for the claim???

• Do you mean maximal value? Or relative maximum? By definition there can be only one maximum. Apr 21, 2015 at 4:50

We have $f(0)=f(1)=0$ and $f(x)$ is positive for $0\lt x\lt 1$.

Thus there is a local (and global) maximum in the interval $(0,1)$.

To show there cannot be two or more local maxima, note that (1) $f'(x)=0$ at local maxima, and (2) if we have a local maximum at $a$ and $b$, then $f'(x)=0$ somewhere between $a$ and $b$. But calculation shows that the first derivative is something harmless times a quadratic, so cannot be $0$ at $3$ places in $(0,1)$.

Remark: Computing the second derivative can be a strategic error, for second derivatives are often a mess. Analysis of the first derivative is usually more informative.

The proposed method will not work, as $f''(x)$ is not always negative; for example try $p=3$.
However, all is not lost. Taking just one derivative: $$f'(x)=\frac{x^{p-1}}{(2-x)^2}(px^2+(-3p-1)x+2p)$$
Note that $\frac{x^{p-1}}{(2-x)^2}>0$ for all $p>1$, $x\in(0,1)$. The second bit is an upward-facing parabola. It is positive for $x=0$ and negative for $x=1$. The vertex is at $x=(3+\frac{1}{p})/2>\frac{3}{2}$, after which the parabola curves upward again. Hence there is exactly one value of $x\in(0,1)$ where $f'(x)$ switches from positive to negative, which is the local max you seek.