Is this Solution Correct. Simple Analysis Question Bounded Function From Below.

Question

For constants $a_1,a_2>0$, $r\in(0,1)$ the function $f:\mathbb{R}_+\to \mathbb{R}$

$$ f(x) =a_1 x-a_2(1-x)^r $$ is bounded from below.

Is the following solution correct? a simple yes is enough $\underline{Solution :}$

Since $f$ is continuous its bounded below on any interval $[0,c],~c<\infty$. $f'(x)=a_1-ra_2(1-x)^{r-1}$ which is positive for some $x$ large enough (depending on $a_1,a_2,r$).

Answer 1

I don't really understand why you calculate $f'$ since you only want to minorate $f$.

As you said $f$ is minorated on any interval $[0, c]$ (the image of a segment by a continuous function is a segment).

Since $a_1 >0$, we have $f(x)\rightarrow +\infty$ when $x\rightarrow +\infty$. Let's take $c$ such that $f(x) > 0$ for all $x>c$.

Then $f$ is minorated on $[0, c]$, and minorated by $0$ on $[c,+\infty]$.

So $f$ is minorated.