Question

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

• A. $[\sqrt{3},\sqrt{7}]$
• B. $[\sqrt{3},\sqrt{5}]$
• C. $[\sqrt{2},\sqrt{3}]$
• D. $[\sqrt{3},\sqrt{6}]$

Answer 1

Answer: (D)

$[\sqrt{3},\sqrt{6}]$

Answer 2

To find the range of the function $f(x) = \sqrt{4-x^2} + \sqrt{x^2-1}$, we follow these steps:

1. Determine the domain:
   We need $4-x^2 \ge 0$ and $x^2-1 \ge 0$. This implies $-2 \le x \le 2$ and ($x \le -1$ or $x \ge 1$). Thus, the domain is $[-2, -1] \cup [1, 2]$.

2. Substitute $u = x^2$:
   As $x$ varies over the domain, $u$ varies over $[1, 4]$. The function becomes $g(u) = \sqrt{4-u} + \sqrt{u-1}$.

3. Find the range of $g(u)$:
   We analyze $g(u)$ on the interval $[1, 4]$.

   • Evaluate at endpoints: $g(1) = \sqrt{3}$ and $g(4) = \sqrt{3}$.
   • Find critical points: $g'(u) = \frac{-1}{2\sqrt{4-u}} + \frac{1}{2\sqrt{u-1}}$. Setting $g'(u) = 0$, we get $u = \frac{5}{2}$.
   • Evaluate at the critical point: $g\left(\frac{5}{2}\right) = \sqrt{4 - \frac{5}{2}} + \sqrt{\frac{5}{2} - 1} = \sqrt{\frac{3}{2}} + \sqrt{\frac{3}{2}} = 2\sqrt{\frac{3}{2}} = \sqrt{6}$.

4. Compare values:
   We have $g(1) = \sqrt{3}$, $g(4) = \sqrt{3}$, and $g\left(\frac{5}{2}\right) = \sqrt{6}$.

5. Determine the range:
   The minimum value is $\sqrt{3}$ and the maximum value is $\sqrt{6}$. Therefore, the range of $f(x)$ is $[\sqrt{3}, \sqrt{6}]$.