Let f:R−35→R be defined by f(x)=3x+25x−3. Then

f(x)=3x+25x−3Let f(x)=y=3x+25x−3⇒5xy−3y=3x+2⇒x(5y−3)=2+3y⇒x=2+3y5y−3∴f−1(x)=2+3x5x−3=f(x)

Let f:R−35→R be defined by f(x)=3x+25x−3. Then

f(x)=3x+25x−3Let f(x)=y=3x+25x−3⇒5xy−3y=3x+2⇒x(5y−3)=2+3y⇒x=2+3y5y−3∴f−1(x)=2+3x5x−3=f(x)

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Let $f : R - \{\frac{3}{5}\} \to R$ be defined by $f (x) = \frac{3 x + 2}{5 x - 3}$ . Then

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$\begin{aligned} f^{- 1} (x) = f (x) \end{aligned}$

$\begin{aligned} f^{- 1} (x) = - f (x) \end{aligned}$

$(fof) x = - x$

$f^{- 1} (x) = \frac{1}{19} f (x)$

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$f (x) = \frac{3 x + 2}{5 x - 3}$

Let $f (x) = y = \frac{3 x + 2}{5 x - 3}$

$\begin{array}{l} \Rightarrow 5 x y - 3 y = 3 x + 2 \Rightarrow x (5 y - 3) = 2 + 3 y \\ \Rightarrow x = \frac{2 + 3 y}{5 y - 3} \\ ∴ f^{- 1} (x) = \frac{2 + 3 x}{5 x - 3} = f (x) \end{array}$

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Let f:R−35→R be defined by f(x)=3x+25x−3. Then

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Let $f : R - \{\frac{3}{5}\} \to R$ be defined by $f (x) = \frac{3 x + 2}{5 x - 3}$ . Then