Variation of $f(x)=x^2-\sqrt{x}$ over $(0;\infty)$ without using derivative (For student didn't get derivation method)?

Question

Am looking to study varition of $f(x)=x^2-\sqrt{x}$ over $(0;\infty)$ without using derivative sign or anything related to derivative function, I want for example using inequalities . We have $f(0)=0$ and $f(1)=0$ then we have f is increasing and decreasing in $(0,1]$ then we must subdivise $(0,1]$ in $(0,a]$,$[a,1], a\in \mathbb{R}$ , for $[a,\infty[$ we have :$f(a) < f(+\infty) $ its little clear that is increasing in that domain , But how I can show its variation for student didn't get derivation method over $(0;\infty)$?

Note: Student didn't get inverse function definition

Answer 1

By AM-GM $$x^2-\sqrt{x}=x^2+\frac{3}{\sqrt[3]{256}}-\sqrt{x}-\frac{3}{\sqrt[3]{256}}\geq$$ $$\geq4\sqrt[4]{x^2\left(\frac{1}{\sqrt[3]{256}}\right)^3}-\sqrt{x}-\frac{3}{\sqrt[3]{256}}=-\frac{3}{\sqrt[3]{256}}.$$ Can you end it now?

Answer 2

We may start with (for $x>0$) : $\displaystyle\;f(x^2)=x^4-x=(x^3-1)\,x\,$

resolve the case $x=1$ and use inequalities for the remaining cases :

• for $x>1$ we have $\;x^3>1\,$ so that $\;(x^3-1)>0$.
  We thus have the product of two increasing positive functions that will also be increasing since $\;\displaystyle f(x+h)g(x+h)-f(x)g(x)=[f(x+h)g(x+h)-f(x)g(x+h)]+[f(x)g(x+h)-f(x)g(x)]$
• for $0<x<1$ we have $\;0<x^3<1\;$ and things are more complicated since $\,1-x^3\,$ is decreasing while $\,x\,$ is increasing. The function $\,x\mapsto x^4-x\,$ goes from $\,0\,$ to $\,0\,$ in $\,(0,1)\,$ so let's see if it may be rewritten as $\,P(x)(x-k)^2+c\,$ with $P(x)$ a polynomial.
  The idea is that the tangent at $\,x=k\,$ will be horizontal with ordinate equal to $\,c\,$ (this is clear in general since its (forbidden) derivative will have a factor $(x-k)\;$).
  $P(x):=x^2+ax+b\,$ (what else?) gives us : \begin{align} x^4-x&=(x^2+ax+b)(x-k)^2+c\\ 0&=(a-2k)x^3+(b+k^2-2ak)x^2+(ak^2-2bk+1)x+c+bk^2\\ a&=2k,\,b=3k^2,\,c=-3k^4,\;k^3=\dfrac 14\quad(\text{from}\ \ 2k^3-6k^3+1=0) \\ f(x^2)&=((x+k)^2+2k^2)(x-k)^2-3k^4\quad\text{with}\ \ k=\dfrac 1{\sqrt[3]4}\\ \end{align} I'll let you conclude.