What is the range of the function $\sqrt{16 - x^{4}}$ $sqrt(16-x^4)$ ?

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Answer 1 · 2017-09-11T15:16:32

See below.

The minimum value $(16 - x^{4})$ $(16 - x^4)$ is $0$ $0$ for real numbers.

Since $x^{4}$ $x^4$ is always positive maximum value of radicand is $16$ $16$

If including both positive and negative outputs the range is:

$[- 4, 4]$ $[ -4 , 4 ]$

For positive output $[0, 4]$ $[ 0 , 4 ]$
For negative output $[- 4, 0]$ $[ -4 , 0 ]$

Theoretically ' $f (x) = \sqrt{16 - x^{4}}$ $f(x) = sqrt( 16- x^4)$ is only a function for either positive or negative outputs, not for both.i.e:

$f (x) = \pm \sqrt{16 - x^{4}}$ $f(x) = +- sqrt(16 - x^4 )$ is not a function.

What is the range of the function sqrt(16-x^4)√16−x4sqrt(16-x^4)?