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

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

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Answer 1 · 2018-04-11T01:10:57

$g (3) = 6$ $g(3) =6$

Explanation:

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

$g (3) = \sqrt[3]{3^{2} - 1} + 2 \sqrt{3 + 1}$ $g(3)=root(3)(3^2-1) + 2sqrt(3+1)$
$g (3) = \sqrt[3]{8} + 2 \sqrt{4}$ $g(3) = root(3)8 + 2sqrt4$
$g (3) = 2 + 2 \sqrt{4}$ $g(3) = 2 + 2sqrt4$
$g (3) = 2 + 2 \times 2$ $g(3) = 2 + 2xx2$
$g (3) = 2 + 4$ $g(3) = 2 + 4$
$g (3) = 6$ $g(3) =6$

Answer 2 · 2018-04-11T01:14:02

$g (3) = 6$ $g(3)=6$

Explanation:

Plug $3$ $3$ in for $x$ $x$

$g (3) = \sqrt[3]{3^{2} - 1} + 2 \sqrt{3 + 1}$ $g(3)=root3(3^2-1)+2sqrt(3+1)$

Simplify

$g (3) = \sqrt[3]{9 - 1} + 2 \sqrt{4}$ $g(3)=root3(9-1)+2sqrt(4)$
$g (3) = \sqrt[3]{8} + 2 \sqrt{4}$ $g(3)=root3(8)+2sqrt(4)$
$g (3) = 2 + 2 (2)$ $g(3)=2+2(2)$
$g (3) = 2 + 4$ $g(3)=2+4$
$g (3) = 6$ $g(3)=6$

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

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Explanation:

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If g(x)=3√x2−1+2√x+1g ( x ) = \root [ 3] { x ^ { 2} - 1} + 2\sqrt { x + 1} , what is g(3)g(3)?

2 Answers

Explanation:

Explanation:

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