1 Answer

Demonchand · Answer 1 · 2022-05-15T12:09:05+0000

Given the function,

$t(v) = \sqrt{ {v}^(2) + 2v - 8 }$

We are aware that square root function is defined for only non negative values.

So,

$\to {v}^(2) + 2v - 8 \geqslant 0$

$\to{v}^(2) + 4v - 2v- 8 \geqslant 0$

$\to {v}(v + 4) - 2(v + 4) \geqslant 0$

$\to(v + 4)(v - 2) \geqslant 0$

Therefore,

$\boxed{\red{v\in (-\infty, -4] \cup [2,\infty})}$

Since this is the possible input values of v for which the given function is defined, it is the domain of the function.

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