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We are asked to find the range of the function$$f(x)=\sqrt{x-3}+\sqrt{12-3x}$$ range

Following is solution $$\sqrt{x-3}=\sin{t},\sqrt{4-x}=\cos{t},0\le t\le\dfrac{\pi}{2}$$it easy to find this $$f=\sin{t}+\sqrt{3}\cos{t}=2\sin{(t+\frac{\pi}{3})}\in [1,2]$$

Is there any other quicker or smarter way to find the range of this kind of functions? (without calculus?)

Many thanks for any help!

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Use Cauchy–Schwarz inequality $$1\cdot\sqrt{x-3}+\sqrt3\sqrt{4-x}\le\sqrt{1+3}\cdot\sqrt{x-3+4-x}=2$$

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  • $\begingroup$ Thanks , But this lower bound we can't use inequality to find it? $\endgroup$ Nov 7 '16 at 10:00
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First, the function's domain is:

$$\begin{cases}x-3\ge0\implies x\ge3\\{}\\12-3x\ge0\implies x\le4\end{cases}\;\;\implies\;\;\;3\le x\le4$$

Now, we can do derivatives:

$$f'(x)=\frac1{2\sqrt{x-3}}-\frac3{2\sqrt{12-3x}}=\frac12\frac{\sqrt{12-3x}-3\sqrt{x-3}}{\sqrt{x-3}\sqrt{12-3x}}$$

The denominator is always non negative, as for the numerator we get (inside our domain!):

$$\sqrt{12-3x}-3\sqrt{x-3}\ge0\iff12-3x\ge9x-27\iff12x\le39\iff x\le\frac{13}4$$

and thus we get our function's derivative fulfills $\;f'(x)\ge0\;$ for all $\;3\le x\le\frac{13}4\;$ and thus it is increasing here and decreasing in $\;\frac{13}4\le x\le4\;$ , and since

$$\begin{cases}f(3)=\sqrt3\\{}\\f\left(\frac{13}4\right)=\frac12+\frac32=2\\{}\\f(4)=1\end{cases}$$

we can see the range is $\;[1,2]\;$ as the function's continuous (with $\;\left(\frac{13}4,\,2\right)\;$ being a local maximum)

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  • $\begingroup$ "we can see the domain is" -> you mean range? $\endgroup$
    – StackTD
    Nov 7 '16 at 10:25
  • $\begingroup$ @StackTD Yes, of course. Good catch, thanks. $\endgroup$
    – DonAntonio
    Nov 7 '16 at 10:40
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It is clear that the domain of $f$ is the interval $[3,4]$. Besides $f$ has a maximun in $x$ the solution of $f\space'(x)=0$ which is $x=\dfrac{39}{12}$. This maximun is equal to $2$. It follows the range is $[1,2]$ as you have found because the minimun $1$ clearly corresponds to $x=4$.

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