Question

$S_1:$ If $f\left(x\right)$ is an increasing function with upward concavity, then concavity of $f^{-1}\left(x\right)$ is also upwards.
$S_2:$ If $f\left(x\right)$ is decreasing function with upwards concavity, then concavity of $f^{-1}\left(x\right)$ is also upwards.

• A. $S_1$ and $S_2$ both are true
• B. $S_1$ is true and $S_2$ is false
• C. $S_1$ is false and $S_2$ is true
• D. $S_1$ and $S_2$ both are false

Answer 1

The correct option is C $S_1$ is false and $S_2$ is true

Let $g\left(x\right)$ be the inverse function of $f\left(x\right).$
Then,
$f\left(g\right(x\left)\right)=x$
$f^{\prime }\left(g\right(x\left)\right)\cdot g^{\prime }\left(x\right)=1$
$i.e.g^{\prime }\left(x\right)=\frac{1}{f^{\prime }\left(g\right(x\left)\right)}$
$\therefore g^{\prime \prime }=-\frac{1}{\left(f^{\prime }\right(g\left(x\right)){)}^2}\cdot f^{\prime \prime }\left(g\right(x\left)\right)\cdot g^{\prime }\left(x\right)$
In $S_1:\text{upward concave},f^{\prime \prime }>0$
$\Rightarrow f^{\prime \prime }\left(g\right(x\left)\right)>0$
and $g^{\prime }\left(x\right)>0\left(f\right(x\left)\text{is increasing so}g\right(x\left)\right)$
$\Rightarrow g^{\prime \prime }\left(x\right)<0$
$\Rightarrow$ Concavity of $f^{-1}\left(x\right)$ is downwards.
$\therefore S_1$ is False
In $S_2:f^{\prime \prime }\left(g\right(x\left)\right)>0$
and $g^{\prime }\left(x\right)<0\left(f\right(x\left)\text{is decreasing so}g\right(x\left)\right)$
$\Rightarrow g^{\prime \prime }\left(x\right)>0$
$\Rightarrow$ Concavity of $f^{-1}\left(x\right)$ is upwards.
$\therefore S_2$ is True.