Question

If $g$ is inverse of $f$ and $f\left(x\right)=x^2+3x-3(x>0)$ then $g^{{}^{\prime }}\left(1\right)$ equals

• A. $\frac{1}{2g\left(1\right)+3}$
• B. $-1$
• C. $\frac{1}{5}$
• D. $\frac{-f^{{}^{\prime }}\left(1\right)}{{\left(f\left(1\right)\right)}^2}$

Answer 1

Answer: (A), (C)

$\frac{1}{2g\left(1\right)+3}$
$\frac{1}{5}$

Given, $f\left(x\right)=x^2+3x-3$

$f^{-1}\left(x\right)=g\left(x\right)$

$\Rightarrow x=f\left(g\left(x\right)\right)$

Differentiating both sides,

$1=f^{{}^{\prime }}\left(g\left(x\right)\right)g^{{}^{\prime }}\left(x\right)$

$\Rightarrow g^{{}^{\prime }}\left(x\right)=\frac{1}{f^{{}^{\prime }}\left(g\left(x\right)\right)}$

Now $f^{{}^{\prime }}\left(x\right)=2x+3$

And $f\left(1\right)=1\Rightarrow g^{\prime }\left(1\right)=\frac{1}{f^{\prime }\left(g\right(1\left)\right)}=\frac{1}{f^{\prime }\left(1\right)}=\frac{1}{5}$