Find the range of $f(x) =(x-1)^{1/2} + 2\cdot(3-x)^{1/2}$ How to take out the range of the following function : 
$$f(x) =(x-1)^{1/2} + 2\cdot(3-x)^{1/2}$$
I am new to functions hence couldn't come up with a solution.
 A: Without Calculus I don't really know how efficiently to find the range of the function. (see edit below)
As far as "naïve" goes, one can find the maximum by Cauchy-Schwarz
$$(\sqrt{x-1}+2\sqrt{3-x})^2\le (1^2+2^2)(x-1+3-x)=10,$$
obtained (only) at $x$ such that $4(x-1)=3-x$, i.e., $x=\frac75$.
And the minimum
$$(\sqrt{x-1}+2\sqrt{3-x})^2\ge (x-1)+4(3-x)\ge11-3x\ge 2,$$
obtained (only) at $x=3$.
However, one needs the Intermediate Value Theorem to ensure that $f(x)$, being continuous on $[1,3]$, attains all values in $[\sqrt{2},\sqrt{10}]$.
Edit: Of course one can pay a little more effort and recover the range using the continuity of square function as follows.
Let $t=\sqrt{3-x}$, then $0\le t\le \sqrt 2$. We want to show that for any $a\in[\sqrt 2,\sqrt{10}]$, the following equation has a solution in $[0,\sqrt{2}]$
$$a=2t+\sqrt{2-t^2}.$$
Equivalently
$$\begin{aligned}a-2t&=\sqrt{2-t^2}\\
(a-2t)^2&=2-t^2\\
5t^2-4at+(a^2-2)&=0.
\end{aligned}
$$
The discriminant is
$$4a^2-5(a^2-2)=10-a^2\ge 0$$
which shows that the last equation has a solution. 
It's easy to see that both solutions are non-negative since both $4a$ and $a^2-2$ are. It remains to show that at least one solution is $\le \sqrt{2}$. It's easy and left as an exercise (hint: $\color{white}{\text{what is their product}}$).
A: As  $1\le x\le 3\iff-1\le x-2\le1$ which nicely fits with the range of sinusoidal functions,
let $x-2=\cos2\theta$ 
So,  $f(x)=\sqrt{2\cos^2\theta}+2\sqrt{2\sin^2\theta}=\sqrt2(|\cos\theta|+\sqrt2|\sin\theta|)$
Case $\#1:$
If $\displaystyle0\le\theta\le\dfrac\pi2,$
$\displaystyle f(x)=\sqrt2[\cos\theta+ 2\sin\theta]=\sqrt2\sqrt5\cos\left(\theta-\arctan2\right)$
As $\displaystyle0\le\theta\le\dfrac\pi2,0-\arctan2\le\theta-\arctan2\le\dfrac\pi2-\arctan2$
Case $\#1A:$
For $\displaystyle-\arctan2\le\theta-\arctan2\le0,$
as $\cos(x)$ is increasing function in $\left[-\dfrac\pi2,0\right];$
$\displaystyle\cos\left(-\arctan2\right)\le\cos\left(\theta-\arctan2\right)\le\cos0$
i.e.,$\displaystyle\frac1{\sqrt5}\le\cos\left(\theta-\arctan2\right)\le1 \implies\sqrt{10}\frac1{\sqrt5}\le f(x)\le\sqrt{10}$
So, $f(x)$ will attain its maximum when $\displaystyle\theta-\arctan2=0$
$\displaystyle\implies\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}=\frac{1-4}{1+4}=-\frac35\implies x=2+\left(-\frac35\right)$
Similarly, $f(x)$ will attain its minimum value i.e., $\displaystyle\sqrt{10}\cdot\frac1{\sqrt5}=\sqrt2$ when $\displaystyle\theta-\arctan2=-\arctan2\iff\theta=0\implies x=2+\cos(2\cdot0)=3$
Case $\#1B:$
For $\displaystyle0\le\theta-\arctan2\le\dfrac\pi2-\arctan2,$
as $\cos(x)$ is decreasing function in $\left[0,\dfrac\pi2\right];$
$\displaystyle\cos0\ge\cos\left(\theta-\arctan2\right)\ge\cos\left(\dfrac\pi2-\arctan2\right)$
i.e., $\displaystyle1\ge\cos\left(\theta-\arctan2\right)\ge\sin\left(\arctan2\right)=\frac2{\sqrt5}$
$\displaystyle\implies\sqrt{10}\ge f(x)\ge\sqrt{10}\frac2{\sqrt5}=2\sqrt2$ which is greater than the earlier minimum $\sqrt2$
I leave for you to deal with
Case $\displaystyle\#2:\frac\pi2<\theta\le\pi$  
Case $\displaystyle\#3:\pi<\theta\le\frac{3\pi}2$  
Case $\displaystyle\#4:\frac{3\pi}2<\theta\le2\pi$  
to show that $\sqrt2\le f(x)\le\sqrt{10}$
A: Your function
$$f(x) =\sqrt{x-1} + 2\sqrt{3-x}$$
is defined for all $x \in [1, 3]$, otherwise the square root are not defined for real number. 
We can check what happens on the border of the segment $[1, 3]$:


*

*$f(1) = 2\sqrt{2}$;

*$f(3) = \sqrt{2}$.


Also, we can find the local maxima and minima of $x$ in the segment $[1, 3]$. In order to get them, let's evaluate the first derivative of $f(x)$:
$$\frac{\text{d} f(x)}{\text{d} x} = -\frac{1}{2\sqrt{x-1}} + \frac{1}{\sqrt{3-x}}$$
and find the values of $x$ such that $\frac{\text{d} f(x)}{\text{d} x} = 0$. Thus:
$$-\frac{1}{2\sqrt{x-1}} + \frac{1}{\sqrt{3-x}} = 0 \Rightarrow 2\sqrt{x-1}=\sqrt{3-x}.$$
Since both $x-1$ and $3-x$ are non-negative in the segment $[1, 3]$, we can elevate to the power of $2$ both members without particular problems:
$$4(x-1)=3-x \Rightarrow 4x+x = 3 + 4 \Rightarrow x = \frac{7}{5} \in [1, 3].$$
$x = \frac{7}{5}$ is a local maximum or minimum and $f\left(\frac{7}{5}\right) = \sqrt{10}.$
Summarizing:


*

*At the begining of the segment $[1, 3]$ the function assumes value $2\sqrt{2}$;

*at the end, the values is $2\sqrt{2}$;

*somewhere inside the segment (i.e. when $x = \frac{7}{5}$) the function has value $\sqrt{10}$.


It's clear now that the minimum value of $f(x)$ in $[1,3]$ is $\sqrt{2}$ while the maximum is $\sqrt{10}$.
We can conclude that the range of $f(x)$ is $[\sqrt{2}, \sqrt{10}]$. 
A: Let $y = f(x)$ be a function with domain $D$.
Let $\left[a,b\right] \le D$, then the global extrema are the extreme (largest and smallest) values in $\left[a,b\right]$.
There are two global extrema; global minima (least value) and global maxima (largest value)
Conceptually speaking,
$$f'(x) = 0\iff f(x)  \text{ `makes a turning`}\iff f(x) \text{ is an extrema}$$
Let $c_1,\space c_2,\space \dots, c_n$ be the points at which $f'(x) = 0$ (ie, the extreme points)
Then,
$$\text{Global Minima }(m) = \min\{f(a), f(c_1),\dots, f(c_n), f(b)\}\\
\text{Global Maxima }(M) = \max\{f(a), f(c_1),\dots, f(c_n), f(b)\}$$
Then, the range of $f(x)$ in $\left[a,b\right]$ is simply the interval $(m,M)$

Given :: $f(x) = \sqrt{x-1} + 2\sqrt{3-x}$

Since $\sqrt{n}$ is only a real function when $n\ge 0$, then in $f(x)$,
$$x-1\ge 0 \quad \&\quad 3-x \ge 0\\
\Rightarrow 1 \le x \le 3\\
\Rightarrow x \in \left[1,3\right]$$
Now, to check for extrema,
$$\frac{d}{dx} f(x) = f'(x) = 0 \\
\Rightarrow \frac{1}{2\sqrt{x-1}} - \frac{1}{\sqrt{3-x}} = 0\\
\Rightarrow \frac1{2\sqrt{x-1}} = \frac1{\sqrt{3-x}}\\
\Rightarrow 4(x-1) = 3-x \\
\Rightarrow x = \frac{7}{5} = 1.4 \in \left[1,3\right]
$$
The suspects:
$$
f\left(\frac{7}{5}\right) = \sqrt{\frac{7 - 5}{5}} + 2\sqrt{\frac{15-7}{5}} = \sqrt{\frac{2}{5}} + 4\sqrt\frac{2}{5} = \sqrt5\cdot\sqrt2 =\sqrt{10} \approx 3.1622\\
f(1) = 0 + 2\sqrt{3-1} = 2\sqrt{2} \approx 2.8284\\
f(3) = \sqrt{3-1} + 0 = \sqrt{2}\approx 1.4142
$$
Clearly, $f\left(\frac{7}{5}\right) > f(1) > f(3)$
$\therefore $ The global Maxima is $\sqrt{10}$ and the global minima is $\sqrt{2}$
$$\boxed{\text{Range}(f) = \left[ \sqrt{2}, \sqrt{10} \right]}$$
A: Let's take a look at the graph of this function:  
 
Before we can find the range, it is helpful to find the domain first. It's pretty clear from the graph that it is defined from $x = 1$ to $x = 3$, or the interval $[1,3]$. To be sure, you can determine it algebraically by finding where both of the square roots are defined; that is, by solving $x - 1 \geq 0$ and $3 - x \geq 0$ and taking the interval where both are true.  
The range is the interval of y-values the function takes on. Usually this is determined using calculus, but we're going to have to use different methods to determine it. In particular, we'll find the x-coordinates of the highest and lowest points of the function (also known as global extrema) from the graph, and then we'll plug them into the function to get the y-values.  
Looking at the graph, we can see that the lowest point of the graph is at the upper end of the domain at $x = 3$. Plugging it into the function, we get:
$$y_{min} = \sqrt{3 - 1} + 2\sqrt{3 - 3} \implies y_{min} = \sqrt{2}$$  
As for the highest point, we can't tell exactly the x-coordinate from the graph, but we can approximate it's value. We see that the function plateaus around $x = 1.4$, so we take this value and plug it into the function:
$$y_{max} = \sqrt{1.4 - 1} + 2\sqrt{3 - 1.4} \implies y_{max} = \sqrt{0.4} + 2\sqrt{1.6} = \sqrt{10}$$
  So the range is the interval $[\sqrt{2}, \sqrt{10}]$.  
It is important to note that in this case we happened to stumble upon the exact x-value for the upper value since we have a computer-generated graph at hand, but this won't always be the case, and even then it still is hard to guess the exact value. Say it was at $x = 1.4001$ instead of $x = 1.4$, then we would have never been able to differentiate between the two values. The best approach is to use calculus and find and test the critical points, as @Nick did a good job of explaining. This is only a way to approximate the range.  
A: The fact that this function is "concave" (i.e. looks like a frowny face) can tell us all we need to know in order to find the range; a property of concave function that is useful for finding the maximum is that any concave function can be bounded above by lines through each point of the graph. We can easily derive a linear upper bound for the square root function. If we start with the inequality of two values $x$ and $a$ chosen to make both sides positive:
$$x+a\leq x+a+\frac{a^2}{4x}$$
which is clearly true, we can factor the left side to
$$x+a\leq (\sqrt{x}+\frac{a}{2\sqrt{x}})^2$$
and take the square root thereof
$$\sqrt{x+a}\leq \sqrt{x}+\frac{a}{2\sqrt{x}}$$
giving a linear upper bound on the square root - that is, if we take $x$ as a constant, the above inequality tells us that as we increase or decrease the argument by changing $a$, the square root must remain below a line. Since your function $f$ is composed entirely out of square roots, we can use this to find an upper bound for it. In particular, if we fix an $x$, notice that
$$\sqrt{x+a-1}\leq \sqrt{x-1}+\frac{a}{2\sqrt{x-1}}$$
and
$$2\sqrt{3-x-a}\leq 2\sqrt{3-x}-\frac{a}{\sqrt{3-x}}.$$
Thus, we can write the bound, by summing the above two inequalities:
$$f(x+a)\leq f(x) + \left(\frac{1}{2\sqrt{x-1}}-\frac{1}{\sqrt{3-x}}\right)a$$
which, at each point $x$, gives a line, parametrized in $a$, as an upper bound to $f$. However, if the coefficient of $a$ were $0$, this would mean the bound would read $f(x+a)\leq f(x)$, meaning that $f(x)$ would be at least as large as any other point $f(x+a)$. We solve for this algebraically by searching for a solution, in $x$, to where coefficient of $a$ is zero in the above equation; that is:
$$\frac{1}{2\sqrt{x-1}}-\frac{1}{\sqrt{3-x}}=0$$
$$\frac{1}{2\sqrt{x-1}}=\frac{1}{\sqrt{3-x}}$$
Reciprocating and squaring both sides yields
$$4x-4=3-x$$
$$5x=7$$
telling us that at $x=\frac{7}{5}$ we have the inequality
$$f\left(\frac{7}5+a\right)\leq f\left(\frac{7}5\right)$$
meaning that $f$ obtains its maximum at $f\left(\frac{7}5\right)=\sqrt{10}$.
To establish the minimum, we use a different property of concave functions (which is often taken to be the definition). In particular, we have, for any $x,y$ and $0\leq\alpha\leq 1$, that the following condition holds:
$$\sqrt{\alpha x+ (1-\alpha)y}\geq \alpha\sqrt{x}+(1-\alpha)\sqrt{y}$$
or, in intuitive terms, if you take two points on the square root function, and draw a line segment between them, this line segment will always be below the graph. You could prove that this is true by squaring both sides and using the AM-GM inequality*. It should be clear that, since $f$ is a sum of shifts and reflections (which preserve concavity) of the square root function, it still holds that
$$f(\alpha x + (1-\alpha)y)\geq \alpha f(x)+(1-\alpha)f(y)$$
but, noting that $f$ has a range of $[1,3]$, if we take $x=1$ and $y=3$, we get:
$$f(3-2\alpha)\geq \alpha f(1) + (1-\alpha)f(3) \geq \min(f(1),f(3))$$
the last inequality holding since $\alpha f(1) + (1-\alpha)f(3)$ is a weighted average of the two and can never be less their minimum. Note that, since $3-2\alpha$ runs through the entire domain of $f$, it must be that for any $x$, we have
$$f(x)\geq \min(f(1),f(3)).$$
So, we just compute $f(1)=2\sqrt{2}$ and $f(3)=\sqrt{2}$, noting that the latter is lesser, so we have
$$f(x)\geq \sqrt{2}$$
and, bringing in our previous result for the maximum, and noting that $f$ is continuous and we have shown that the extrema are obtained, we find the range of $f$ to be $$[\sqrt{2},\sqrt{10}]$$
You could find the extrema of any function of the form $f(x)=a\sqrt{x-b}+c\sqrt{d-x}$ for non-negative $a$ and $c$ using exactly my technique, where minimum lies at either $x=b$ or $x=d$ and the maximum is at the solution, in $x$, to $\frac{a}{2\sqrt{x-b}}-\frac{c}{2\sqrt{d-x}}=0$, which occurs at $x=\frac{bc^2+ad^2}{a^2+c^2}$. You could actually keep going to add more terms of the form $\alpha\sqrt{x-\beta}$ or $\alpha\sqrt{\beta-x}$ for $\alpha\geq 0$ and still apply basically the same technique. The minima would still be at the edges of domain, however, the equation for the maximum would have more terms of the form $\frac{\alpha}{2\sqrt{x-\beta}}$ and would become increasingly difficult, if not impossible, to solve in closed form.
(*Actually, the fact that $f$ is bounded above by a line through every point is equivalent to concavity, so we could prove it that way too. This entire proof is basically working off $f$ being concave - it's a very useful property when you have it.)
