![\bf \begin{cases} f(x)=\sqrt{x+2}\\\\ g(x)=x^3-2 \end{cases}\qquad (g \circ f)(x)\iff g[\ f(x)\ ] \\\\\\ g[\ f(x)\ ]=\left( \boxed{f(x)} \right)^3-2\implies g[\ f(x)\ ]=\left( \boxed{\sqrt{x+2}} \right)^3-2 \\\\\\ g[\ f(x)\ ]=\sqrt{(x+2)^3}-2\implies g[\ f(x)\ ]=(x+2)\sqrt{x+2}-2](/tpl/images/0312/6172/d911d.png)
or x > -3
g(x)=x cubed -2
what is (g open circle f) (x)
Mathematics, 12.10.2019 07:50, hdhdhd49jdhd
F(x) = the square root of x+2
g(x)=x cubed -2
what is (g open circle f) (x)
and the domain?
Answers: 2
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Do you know the correct answer?
F(x) = the square root of x+2
g(x)=x cubed -2
what is (g open circle f) (x)
g(x)=x cubed -2
what is (g open circle f) (x)
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