Range of function $f(x)=\sqrt{4-x^2}+\sqrt{x^2-1}$

Question

Finding range of function $f(x)=\sqrt{4-x^2}+\sqrt{x^2-1}$

What i try

$$y^2=\sqrt{4-x^2}+\sqrt{x^2-1}\Rightarrow y^2=3+\sqrt{(4-x^2)(x^2-1)}$$

$$y^2=3+\sqrt{4x^2-x^4-4+x^2}=3+\sqrt{5x^2-x^4-4}$$

How do i solve it Help me please

Answer 1

Since $\sqrt{}$ is a concave function, by Karamata we obtain: $$\sqrt{4-x^2}+\sqrt{x^2-1}\geq\sqrt{4-x^2+x^2-1}+\sqrt{0}=\sqrt3.$$ The equality occurs for $x^2-1=0,$ which says that we got a minimal value of $f$.

Now, by C-S $$\sqrt{4-x^2}+\sqrt{x^2-1}\leq\sqrt{(1^2+1^2)(4-x^2+x^2-1)}=\sqrt6.$$ The equality occurs for $4-x^2=x^2-1,$ which says that we got a maximal value.

Since $f$ is continuous function on the domain, we got the answer: $$[\sqrt3,\sqrt6].$$

Answer 2

Continuing from where you left off $$y^2=3+2\sqrt{\frac{9}{4}-\left(x^2-\frac{5}{2}\right)^2}$$

Clearly, $$y^2\in \left[3,6\right]$$

$$y\in\left[\sqrt{3},\sqrt{6}\right]$$

Answer 3

$y^2=3+\color{red}2\sqrt{4-x^2}\sqrt{x^2-1}$

Using the AM-GM inequality, $0\le2\sqrt{4-x^2}\sqrt{x^2-1}\le(4-x^2)+(x^2-1)=3.$

Therefore $3\le y^2\le3+3=6.$

Can you take it from here?