How do you find the domain of $f (x) = \sqrt{12 - 2 x^{2}}$ $f(x)=sqrt(12 - 2x^2)$ ?

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Answer 1 · 2018-03-13T16:18:01

$[- \sqrt{6}, \sqrt{6}]$ $[-sqrt(6),sqrt(6)]$

To find the domain of a function, we find all possible values of its input which give a defined function.

Here, we have $\sqrt{12 - 2 x^{2}}$ $sqrt(12-2x^2)$

Since when $x < 0$ $x<0$ the function of $\sqrt{x}$ $sqrt(x)$ will be undefined, we say this function is defined when:

$12 - 2 x^{2} \geq 0$ $12-2x^2>=0$

$6 - x^{2} \geq 0$ $6-x^2>=0$

$- x^{2} \geq - 6$ $-x^2>=-6$

$x^{2} \leq 6$ $x^2<=6$

$- \sqrt{6} \leq x \leq \sqrt{6}$ $-sqrt(6)<=x<=sqrt(6)$

So $f (x)$ $f(x)$ is only defined when it is between $- \sqrt{6}$ $-sqrt(6)$ and $\sqrt{6}$ $sqrt(6)$ . In interval notation, we write this as:

$[- \sqrt{6}, \sqrt{6}]$ $[-sqrt(6),sqrt(6)]$

How do you find the domain of f(x)=√12−2x2f(x)=sqrt(12 - 2x^2)?