Question

If \(f(x)=\dfrac{2}{3}x + \dfrac{3}{2}, x \in R\) then what is f^-1(x) equal to?
1. \(\dfrac{3}{2}x + \dfrac{2}{3}\)
2. \(\dfrac{3}{2}x - \dfrac{9}{4}\)
3. \(\dfrac{2}{3}x - \dfrac{4}{9}\)
4. \(\dfrac{2}{3}x - \dfrac{2}{3}\)

Answer 1

Correct Answer - Option 2 : \(\dfrac{3}{2}x - \dfrac{9}{4}\)

Concept:

Let y = f(x).

To find  f-1(x), use the following steps

(1) Switch the places of x and y

$(2)\; solve\; for\; y$

 

Calculations:

Let y = f(x)

⇒ y = \(\rm \dfrac{2}{3}x + \dfrac{3}{2}\)

Switch the places of  $x$  and  $y$

⇒ x = \(\rm \dfrac{2}{3}y + \dfrac{3}{2}\)

Solve for  $y$

⇒ x - \(\rm \dfrac 32 = \dfrac 2 3 y\)

⇒  \(\rm \dfrac {(x -\dfrac 32)}{\dfrac 23} = y\)

\(⇒ \rm y = \dfrac{3}{2}x - \dfrac{9}{4}\)

Hence, If \(f(x)=\dfrac{2}{3}x + \dfrac{3}{2}, x \in R\)  then  f-1(x) = \(\rm \dfrac{3}{2}x - \dfrac{9}{4}\)