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#