How to find the range of the function: $f(x) = \sqrt{x-1}+2\sqrt{3-x}$

Question

Problem :

Find the range of the function: $f(x) = \sqrt{x-1}+2\sqrt{3-x}$

Solution :

Domain of this function can be determined as :

$x - 1 >0 ; 3-x >0 \Rightarrow x >0 ; x <3 ;$

$\therefore $ domain of $x \in [1,3]$

Now if I put the values of this domain in my function then it gives the following values :

at 1 ; the value of the function is $2\sqrt{2}$

at 2 : the value of the function is $ 1+2 = 3$

at 3 : the value of the function is $2$

Can we say that the maximum value of the function is 3 and minimum value of the function is 2;

Therefore the range of this function is [2,3] but this answer is wrong. please suggest..

Also suggest that how can we use differentiation method to find the range... thanks.

Answer 1

We get $$f^\prime (x)=\frac{1}{2\sqrt{x-1}}-\frac{2}{2\sqrt{3-x}}=\frac{\sqrt{3-x}-2\sqrt{x-1}}{2\sqrt{(x-1)(3-x)}}.$$

So, $$f^\prime(x)\ge0\iff \sqrt{3-x}\ge2\sqrt{x-1}\iff 3-x\ge4(x-1)\iff x\le\frac{7}{5}.$$ Now we know that $f(x)$ is increasing in $1\le x\lt 7/5$ and that $f(x)$ is decreasing in $7/5\lt x\le 3$.

So, we know that the max is $f(7/5)$, and that the min is $\min(f(1),f(3)).$

Answer 2

This function is only defined on $[1,3]$. We can differentiate it, and search for points where the derivative is $0$. Since it is a continuous (on $[1,3]$) and differentiable (on $(1,3)$) function, extreme values must happen either at $1$, $3$ or at such points with derivative $0$.

Answer 3

Since the function is continuous, the image of the closed interval $[1, 3]$, i.e. $f([1, 3])$ is also a closed interval.

Since the function is also differentiable, we can find the maximum and minimum values (which the function attains) by comparing all the critical points (set $f'(x) = 0$) and the end points ($f(1)$ and $f(3)$).

Answer 4

I try to give another solution without using derivative. Here, we only use the intermediate value theorem (https://en.wikipedia.org/wiki/Intermediate_value_theorem) for continuous functions and Cauchy Schwarz inequality (https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality).

Clearly, the domain of the function is [1;3]. For every $x\in[1,3]$, we have \begin{align} f(x)-f(3)&=\sqrt{x-1}-\sqrt{2}+2\sqrt{3-x}\\ &=\frac{x-3}{\sqrt{x-1}+\sqrt{2}}+2\sqrt{3-x}\\ &=\frac{\sqrt{3-x}(2\sqrt{x-1}+2\sqrt{2}-\sqrt{3-x})}{\sqrt{x-1}+\sqrt{2}}\\ &\geq 0. \end{align} By the Cauchy-Schwarz inequality we have \begin{align} f(x)-f(7/5)&\leq \sqrt{(1+4)(x-1+3-x)}-(\sqrt{2}/\sqrt{5}+2\sqrt{8}/\sqrt{5})=0 \end{align} Hence, $f([1,3])\subset [f(3),f(5/7)]$. On the other hand, by the intermediate value theorem, we have $$ f([1,3])\supset f([7/5,3])\supset [f(3),f(7/5)]. $$ The range of $f$ is $[f(3),f(5/7)]$.