Inverse of the Function – Problems

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) = 2x²-x then f⁻¹ (x) =?

Solution: Let y = f (x) = 2x²-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.
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