If $f (x) = 3 x - 5$ then $f^{- 1} (x) =$

\frac{1}{3 x - 5}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{x + 5}{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

does not exist because

f

is not one-one

No worries! We‘ve got your back. Try BYJU‘S free classes today!

does not exist because

f

is not onto

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\frac{x + 5}{3}$
$f (x) = 3 x - 5$

Since $f (x)$ is a purely linear function (straight line), it is therefore a one-one function.

Hence $f (x)$ is revertible for all real $x .$

$y = 3 x - 5$

$y + 5 = 3 x$

$x = \frac{y + 5}{3}$

Replacing $x$ by $y$ , gives us

$f^{- 1} (x) = \frac{x + 5}{3}$

Suggest Corrections

Similar questions

f : R \to R is given by f (x) = 3 x - 5, then f^{- 1} (x)

\frac{1}{3 x - 5}

\frac{x + 5}{3}

f : R - {\frac{3}{5}} \to R - {\frac{3}{5}}

f (x) = \frac{3 x + 1}{5 x - 3}

lim x \to a f (x)

=

{lim}_{x \to a^{-}} f (x) = l a n d {lim}_{x \to a^{+}} f (x) = m

l \neq m .

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} x^{2} + | x |, & x < - 5 25 & x = - 5 \frac{3 x^{2}}{5} + 15 & x > - 5 \end{matrix}

x = - 5

F e I_{3}

For real $x$ , let $f (x) = x^{3} + 5 x + 1$ , then

Join BYJU'S Learning Program