If $g ( x ) = \root [ 3] { x ^ { 2} - 1} + 2\sqrt { x + 1}$ , what is $g(3)$ ?

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Answer 1 · 2018-04-11T01:10:57

$g(3) =6$

Just substitute 3 in wherever there's an $x$

$g(3)=root(3)(3^2-1) + 2sqrt(3+1)$
$g(3) = root(3)8 + 2sqrt4$
$g(3) = 2 + 2sqrt4$
$g(3) = 2 + 2xx2$
$g(3) = 2 + 4$
$g(3) =6$

Answer 2 · 2018-04-11T01:14:02

$g(3)=6$

Plug $3$ in for $x$

$g(3)=root3(3^2-1)+2sqrt(3+1)$

Simplify

$g(3)=root3(9-1)+2sqrt(4)$
$g(3)=root3(8)+2sqrt(4)$
$g(3)=2+2(2)$
$g(3)=2+4$
$g(3)=6$

If g ( x ) = \root [ 3] { x ^ { 2} - 1} + 2\sqrt { x + 1} g ( x ) = \root [ 3] { x ^ { 2} - 1} + 2\sqrt { x + 1} , what is g(3)g(3)?