Question $f(x) = \sqrt{x^{2}-4} + \sqrt{x^{2}+9}$

Studdy Solution

STEP 1

What is this asking? We're trying to find the smallest value of this funky function $f(x) = \sqrt{x^2 - 4} + \sqrt{x^2 + 9}$ . Watch out! We can't just plug in any $x$ value!
We need to make sure what's under the square roots isn't negative.

STEP 2

1. Domain
2. Rewrite the function
3. Minimize the function

STEP 3

Alright, let's figure out what values of $x$ we're allowed to use!
We've got two square roots, so we need to make sure what's underneath them isn't negative.
For the first one, $\sqrt{x^2 - 4}$ , we need $x^2 - 4 \ge 0$ .
Adding $4$ to both sides gives us $x^2 \ge 4$ , which means $x \le -2$ or $x \ge 2$ .

STEP 4

For the second square root, $\sqrt{x^2 + 9}$ , we need $x^2 + 9 \ge 0$ .
Since $x^2$ is always greater than or equal to zero, $x^2 + 9$ will always be greater than zero!
So, we don't have to worry about that one.

STEP 5

Combining our findings, we know $x$ can be any number such that $x \le -2$ or $x \ge 2$ .

STEP 6

Let's think about how we can minimize this function.
Notice that both square roots are always positive.
Also, since $x^2$ shows up in both, larger values of $x^2$ will make the function bigger.

STEP 7

So, the smallest value of the function will likely occur when $x^2$ is as small as possible.
From our domain, the smallest value of $x^2$ is $4$ (when $x = 2$ or $x = -2$ ).

STEP 8

Let's plug in $x=2$ (or $x=-2$ , it doesn't matter since we're squaring it!) into our function: $f(2) = \sqrt{2^2 - 4} + \sqrt{2^2 + 9}$ $f(2) = \sqrt{4 - 4} + \sqrt{4 + 9}$ $f(2) = \sqrt{0} + \sqrt{13}$ $f(2) = 0 + \sqrt{13}$ $f(2) = \sqrt{13}$

STEP 9

The minimum value of the function $f(x)$ is $\sqrt{13}$ , which happens when $x = 2$ or $x = -2$ .

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