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.


4 Answers 4


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 $\min(f(1),f(3)).$


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 $0$.


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 $f(3)$).


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 $[f(3),f(5/7)]$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.