Find the range of the $\sqrt{x-3}+\sqrt{12-3x}$ 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!
 A: Use Cauchy–Schwarz inequality
$$1\cdot\sqrt{x-3}+\sqrt3\sqrt{4-x}\le\sqrt{1+3}\cdot\sqrt{x-3+4-x}=2$$ 
A: 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)
A: 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 $x=4$.
