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Giving some math indications to my brother whom is in high school, I ran into the following:

Sketch $f(x)=\log_2(|x-1|+1)$ and using the graphic solve the following inequality:

$\sqrt{x}+ f(x) \geq 1$.

Now, the graphic is the following (thanks to Desmos):

'where the dotted curve is $\sqrt{x}$'

The problem to me is that at my brother level there is no knowledge of calculus, derivatives and so on, but only elementary (algebraic) properties of some functions,

so how can one solve such a problem using the graphic without calculus?

I mean for $x\geq 1$ it is evident, for $x< 0$ it is undefined, but for $0\leq x<1$ ?

nonuser
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HaroldF
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1 Answers1

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Observe that $\sqrt{x}\geq x$ for all $x\in[0,1],$ and $f(x)\geq 1-x$ for all $x\in[0,1].$ Then $$\sqrt{x}+f(x)\geq x+(1-x)=1\text{ for all }x\in[0,1].$$

Graphically, the two inequalities I used are just observing that these functions lie over the line segments connecting $(0,0)$ and $(1,1)$ (the first inequality) and connecting $(0,1)$ and $(1,0)$ (the second inequality), respectively.

RideTheWavelet
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  • Oh thank you! This is absolutely a solution to the problem! Now I wonder if it is possible to solve $f(x)\geq 1-x$ in the interval without calculus, and this probably reduces to the fact that the exponential is convex, which i think can be done without calculus. E.g. https://math.stackexchange.com/questions/1667300/proof-of-concavity-of-log-function – HaroldF Jan 10 '18 at 22:48