First slide

Introduction to real valued functions

Question

The value of the parameter $' k'$ for which the inverse of $f (x) = 1 + 3 k x, k \neq 0$ is itself is

Easy

A

$\frac{- 1}{3}$
B

$\frac{1}{3}$
C

$- 3$
D

$\frac{- 1}{2}$

Solution

$f^{- 1} (x) = f (x) \Rightarrow \frac{x - 1}{3 k} = 1 + 3 k x$
$\Rightarrow x - 1 = 3 k + 9 k^{2} x$

$\Rightarrow 3 k = - 1 \Rightarrow k = \frac{- 1}{3}$

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