Answer:
2 Function B
Step-by-step explanation:
$5.25, because if you multiply $6 by 7/8(fraction form), you get $5.25
For the given equation;

We shall begin by expanding the parenthesis on the left side, after which we would combine all terms on and move all of them to the left side, which shall yield a quadratic equation. Then we shall solve.
Let us begin by expanding the parenthesis;

Now that we have expanded the left side of the equation, we would have;

We shall now solve the resulting quadratic equation using the quadratic formula as follows;
![\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{Where;} \\ a=9,b=-30,c=150 \\ x=\frac{-(-30)\pm\sqrt[]{(-30)^2-4(9)(150)}}{2(9)} \\ x=\frac{30\pm\sqrt[]{900-5400}}{18} \\ x=\frac{30\pm\sqrt[]{-4500}}{18} \\ x=\frac{30\pm\sqrt[]{-900\times5}}{18} \\ x=\frac{30\pm\sqrt[]{-900}\times\sqrt[]{5}}{18} \\ x=\frac{30\pm30i\sqrt[]{5}}{18} \\ \text{Therefore;} \\ x=\frac{30+30i\sqrt[]{5}}{18},x=\frac{30-30i\sqrt[]{5}}{18} \\ \text{Divide all through by 6, and we'll have;} \\ x=\frac{5+5i\sqrt[]{5}}{3},x=\frac{5-5i\sqrt[]{5}}{3} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%3D%5Cfrac%7B-b%5Cpm%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D%20%5C%5C%20%5Ctext%7BWhere%3B%7D%20%5C%5C%20a%3D9%2Cb%3D-30%2Cc%3D150%20%5C%5C%20x%3D%5Cfrac%7B-%28-30%29%5Cpm%5Csqrt%5B%5D%7B%28-30%29%5E2-4%289%29%28150%29%7D%7D%7B2%289%29%7D%20%5C%5C%20x%3D%5Cfrac%7B30%5Cpm%5Csqrt%5B%5D%7B900-5400%7D%7D%7B18%7D%20%5C%5C%20x%3D%5Cfrac%7B30%5Cpm%5Csqrt%5B%5D%7B-4500%7D%7D%7B18%7D%20%5C%5C%20x%3D%5Cfrac%7B30%5Cpm%5Csqrt%5B%5D%7B-900%5Ctimes5%7D%7D%7B18%7D%20%5C%5C%20x%3D%5Cfrac%7B30%5Cpm%5Csqrt%5B%5D%7B-900%7D%5Ctimes%5Csqrt%5B%5D%7B5%7D%7D%7B18%7D%20%5C%5C%20x%3D%5Cfrac%7B30%5Cpm30i%5Csqrt%5B%5D%7B5%7D%7D%7B18%7D%20%5C%5C%20%5Ctext%7BTherefore%3B%7D%20%5C%5C%20x%3D%5Cfrac%7B30%2B30i%5Csqrt%5B%5D%7B5%7D%7D%7B18%7D%2Cx%3D%5Cfrac%7B30-30i%5Csqrt%5B%5D%7B5%7D%7D%7B18%7D%20%5C%5C%20%5Ctext%7BDivide%20all%20through%20by%206%2C%20and%20we%27ll%20have%3B%7D%20%5C%5C%20x%3D%5Cfrac%7B5%2B5i%5Csqrt%5B%5D%7B5%7D%7D%7B3%7D%2Cx%3D%5Cfrac%7B5-5i%5Csqrt%5B%5D%7B5%7D%7D%7B3%7D%20%5Cend%7Bgathered%7D)
ANSWER:
C^2+(2/3)
2^2+(2/3)
4+(2/3)
4 2/3
Final answer: 4 2/3 or 14/3 as an improper fraction

Actually Welcome to the Concept of the Inverse of real function.
Let's consider here, g(n) = y ,
so we get as,
![y = \sqrt[3]{ \frac{n - 1}{2} }](https://tex.z-dn.net/?f=y%20%3D%20%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7Bn%20-%201%7D%7B2%7D%20%7D%20)
no, cubing the power both side we get as,
=>

now,

so finally, we get as,
=>

hence,here, n = inverse of the g(n) function.
so,
g^-1 (n) = 2y^3+1.