Question

If f(1) = 3, f'(1) = 2 and f"(1) = 4 and let $f^{-1}\left(x\right)=g\left(x\right)$, then g"(3) is equal to

• A. - 2
• B. 2
• C. $-\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

The correct option is C $-\frac{1}{2}$

$f\left(g\right(x\left)\right)=g\left(f\right(x\left)\right)=x$
$g^{{}^{\prime }}\left(f\right(x\left)\right)f^{{}^{\prime }}\left(x\right)=1$
$g^{\prime \prime }\left(f\right(x\left)\right)f^{{}^{\prime }}\left(x\right)=\frac{d}{dx}\left(\frac{1}{f^{{}^{\prime }}\left(x\right)}\right)$
$=\frac{-f^{{}^{\prime \prime }}\left(x\right)}{\left(f^{{}^{\prime }}\right(x){)}^2}$
Put x = 1
$g^{\prime \prime }\left(3\right)=-\frac{1}{2}$