Let gx be the inverse of the function fx and f1(x)=11+x3. Then g1(x) is equal to

# Let $g\left(x\right)$ be the inverse of the function $f\left(x\right)$ and ${f}^{1}\left(x\right)=\frac{1}{1+{x}^{3}}.$ Then ${g}^{1}\left(x\right)$ is equal to

1. A

$\frac{1}{1+{\left(g\left(x\right)\right)}^{3}}$

2. B

$\frac{1}{1+{\left(f\left(x\right)\right)}^{3}}$

3. C

$1+{\left(g\left(x\right)\right)}^{3}$

4. D

$1+{\left(f\left(x\right)\right)}^{3}$

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### Solution:

Since $g\left(x\right)$ is the inverse of $f\left(x\right)$, therefore $f\left(x\right)=y⇔g\left(y\right)=x$

Now,  $⇒{g}^{\mathrm{\prime }}\left(y\right)=1+\left(g\left(y\right){\right)}^{3}$

[using $f\left(x\right)=y⇔x=g\left(y\right)$$⇒{g}^{\mathrm{\prime }}\left(x\right)=1+\left(g\left(x\right){\right)}^{3}$ (replacing y by x)

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