MHB Proving inequality using Mean Value Theorem

[Samme013]

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Need help with this exercise been stuck on it for a while i think i get the gist of what i am supposed to do but can't seem to get it to work i am definitely missing something. I set h(x)= x-2/3 - g(x) and tried using the mean value theorem on [a,b] and then tried finding the minimum value of h'(x) in order to find a relation between h(a),h(b) but an't seem to get the given answer. Thanks in advance , really need to get the exercise done

 

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Need help with this exercise been stuck on it for a while i think i get the gist of what i am supposed to do but can't seem to get it to work i am definitely missing something. I set h(x)= x-2/3 - g(x) and tried using the mean value theorem on [a,b] and then tried finding the minimum value of h'(x) in order to find a relation between h(a),h(b) but an't seem to get the given answer. Thanks in advance , really need to get the exercise done

Hi Samme013,

To prove that the inequality $g(x) \le x^{-2/3}$ is valid for all $x > 0$, apply the mean value theorem to $f$, not $h$. Namely, fix $x > 0$. Since $f$ is continuous on $[x, x+3]$ and differentiable on $(x, x+3)$, the mean value theorem gives a $c\in (x, x+3)$ such that $f(x + 3) - f(x) = f'(c)((x + 3) - 3)$, or $g(x) = 3f'(c)$. Since $f'(c) = \frac{1}{3}c^{-2/3}$, $g(x) = c^{-2/3}$. Since $c > x$, $c^{-2/3} \le x^{-2/3}$. Therefore $g(x) \le x^{-2/3}$. Since $x$ was an arbitrary positive number, $g(x) \le x^{-2/3}$ for all $x > 0$.