Inverse of the Function – Problems

1. The inverse of the function \(\frac{{{10}^{x}}-{{10}^{-x}}}{{{10}^{x}}+{{10}^{-x}}}\) is

Solution: \(y=\frac{{{10}^{x}}-{{10}^{-x}}}{{{10}^{x}}+{{10}^{-x}}}\),

⇒ x = ½ log₁₀ \(\left( \frac{1+y}{1-y} \right)\),

Let y = f (x) ⇒ x = f⁻¹ (y)

⇒ f⁻¹ (y) = ½ log₁₀ \(\left( \frac{1+y}{1-y} \right)\),

⇒ f⁻¹ (x) = ½ log₁₀ \(\left( \frac{1+x}{1-x} \right)\).

2. If f: R → R is defined by f (x) = 3x – 4 then f⁻¹: R → R is

Solution: f (x) = 3x – 4

Let y = f⁻¹ (x)

⇒ f (y) = x

⇒ 3y – 4 = x

⇒ 3y = x + 4

⇒ \(y=\frac{x+4}{3}\),

⇒ f⁻¹ (x) = \(\frac{x+4}{3}\).

3. If the function f: [1, ∞) → [1, ∞) is defined by f (x) = 2^x²-x then f⁻¹ (x) =?

Solution: Let y = f (x) = 2^x²-x,

log₂ y = x² – x ⇒ x² – x – log₂ y = 0,

⇒ \(x=\frac{1\pm \sqrt{1+4{{\log }_{2}}y}}{2}\),

⇒ \(x=\frac{1+\sqrt{1+4{{\log }_{2}}y}}{2}\),

If \(x=\frac{1-\sqrt{1+4{{\log }_{2}}y}}{2}\),

⇒ \(=\frac{1-\sqrt{1+4({{x}^{2}}-x)}}{2}\),

⇒ \(=\frac{1-\sqrt{{{(2x-1)}^{2}}}}{2}\),

⇒ \(=\frac{1-2x+1}{2}\),

= 1 – x,

x = 1 – x,

2x = 1.

x = ½ is not in the domain.