Answer:
D) ![\frac{1}{6}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B6%7D)
g¹(1) = ![\frac{1}{6}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B6%7D)
The inverse of the function ![g(x) = \frac{x^{\frac{1}{3} }-1 }{2}](https://tex.z-dn.net/?f=g%28x%29%20%3D%20%5Cfrac%7Bx%5E%7B%5Cfrac%7B1%7D%7B3%7D%20%7D-1%20%7D%7B2%7D)
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given that f(x) = (2x+1)³
Let y = (2x+1)³
![y^{\frac{1}{3} } =2x+1](https://tex.z-dn.net/?f=y%5E%7B%5Cfrac%7B1%7D%7B3%7D%20%7D%20%3D2x%2B1)
![2x = y^{\frac{1}{3} } -1](https://tex.z-dn.net/?f=2x%20%3D%20y%5E%7B%5Cfrac%7B1%7D%7B3%7D%20%7D%20-1)
![x = \frac{y^{\frac{1}{3} }-1 }{2}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7By%5E%7B%5Cfrac%7B1%7D%7B3%7D%20%7D-1%20%7D%7B2%7D)
<u><em>Step(ii):-</em></u>
y = f(x) ⇒ x = f⁻¹ (y)
⇒ ![f^{-1} (y) = \frac{y^{\frac{1}{3} }-1 }{2}](https://tex.z-dn.net/?f=f%5E%7B-1%7D%20%28y%29%20%3D%20%5Cfrac%7By%5E%7B%5Cfrac%7B1%7D%7B3%7D%20%7D-1%20%7D%7B2%7D)
![f^{-1} (x) = \frac{x^{\frac{1}{3} }-1 }{2}](https://tex.z-dn.net/?f=f%5E%7B-1%7D%20%28x%29%20%3D%20%5Cfrac%7Bx%5E%7B%5Cfrac%7B1%7D%7B3%7D%20%7D-1%20%7D%7B2%7D)
The inverse of the given function
![g(x) = \frac{x^{\frac{1}{3} }-1 }{2}](https://tex.z-dn.net/?f=g%28x%29%20%3D%20%5Cfrac%7Bx%5E%7B%5Cfrac%7B1%7D%7B3%7D%20%7D-1%20%7D%7B2%7D)
Differentiating equation (i) with respective to 'x', we get
![g^{l} (x) = \frac{1}{2} X \frac{1}{3} x^{\frac{1}{3} -1}](https://tex.z-dn.net/?f=g%5E%7Bl%7D%20%28x%29%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20X%20%5Cfrac%7B1%7D%7B3%7D%20x%5E%7B%5Cfrac%7B1%7D%7B3%7D%20-1%7D)
![g^{l} (x) = \frac{1}{6} x^{\frac{-2}{3} }](https://tex.z-dn.net/?f=g%5E%7Bl%7D%20%28x%29%20%3D%20%5Cfrac%7B1%7D%7B6%7D%20%20x%5E%7B%5Cfrac%7B-2%7D%7B3%7D%20%7D)
<u><em>Final answer:-</em></u>
Put x=1
![g^{l} (1) = \frac{1}{6} 1^{\frac{-2}{3} } = \frac{1}{6}](https://tex.z-dn.net/?f=g%5E%7Bl%7D%20%281%29%20%3D%20%5Cfrac%7B1%7D%7B6%7D%20%201%5E%7B%5Cfrac%7B-2%7D%7B3%7D%20%7D%20%3D%20%5Cfrac%7B1%7D%7B6%7D)