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Please don't give the solution, I already got the answer by a different method.

I want to know why the method in the picture is wrong? Why cannot we simple add inequalities like that to get the interval of range and then find the minimum value from that range?

The correct answer to this question is : 4/9 ( minimum value) enter image description here

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  • $\begingroup$ @mfl can you please elaborate on that? How do you prove that? $\endgroup$
    – Arishta
    Dec 22 '16 at 10:24
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You're right that $t\ge9$ and $0<4/t\le 4/9$. Therefore $$ t+\frac{4}{t}-9\ge0 $$ However, this does not tell you that the minimum is $0$, which is not even an attained value. Indeed, you have $t\ge9$ and $4/t>0$, so certainly $$ t+\frac{4}{t}\color{red}{>}9 $$ Thus, just considering those inequalities is not sufficient for determining the minimum value.

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I had a similar question long time back. Why does the range of $\sin x +\cos x$ is [-$\sqrt 2$,$\sqrt 2$] and not [$-2$,$2$] as the range of $\sin x$ and $\cos x$ is [-1,1], where $x$ is real? Basically why can't you add the ranges of two functions to obtain the overall range of their sum (in general). The catch here is that,the two extreme points of their range do not occur for the same value of $x$,in general. Hence you can't add their ranges, as there will be no such $x$ where that value occurs. In your case you can't add the ranges of the functions $t$ and $4/t$ as the minima/maxima for these functions occur for different values of $t$.

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You are right that $t\in [9,\infty)$ and $\frac{4}{t} \in (0,\frac{4}{9})$ (note that it's not $1/9$)

But this deos not imply that $t+4/t \in [9,\infty)$; there are several things wrong with that statement.

First, the expression cannot take value $9$, because the least $t$ can be is $9$, but $4/t$ cannot possibly take value $0$.

Even after assuming that $4/t$ can take $0$, to claim that $t+4/t\ge 9$, there must exist a value $t$ such that $t$ and $4/t$ both take their minimum value $9$ and $0$, which clearly cannot happen.

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  • $\begingroup$ Since $t\ge9$ and $4/t>0$ we can surely state that $t+\frac{4}{t}\ge9$. $\endgroup$
    – egreg
    Dec 22 '16 at 12:03

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