A function f : R → R is defined as f(x) = x3 + 4. Is it a bijecti… - Vidyadip

Question

A function f : R → R is defined as f(x) = x³ + 4. Is it a bijection or not? In case it is a bijection, find f^-1(3).

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Answer

We have,
f : R → R in a function defined by
f(x) = x³ + 4
Injectivity: Let f(x₁) = f(x₂) for $\text{x}_1,\text{x}_2\in\text{R}$
$\Rightarrow\ \text{x}_1^3+4=\text{x}_2^3+4$
$\Rightarrow\ \text{x}_1^3=\text{x}_2^3$
$\Rightarrow\ \text{x}_1=\text{x}_2$
⇒ f is one-one.
Surjectivity: Let $\text{y}\in\text{R}$ be artritrary such that
f(x) = y
⇒ x³ + 4 = y
⇒ x³ + 4 - y = 0
We know that an odd degree equation must have a real root.
$\Rightarrow\ \alpha^3+4=\text{y}$
$\Rightarrow\ \text{f}(\alpha)=\text{y}$
⇒ f is onto.
Since f is one-one and onto.
⇒ f is bijective.
finally, f(x) = y
⇒ x³ + 4 = y
⇒ x³ = y - 4
$\Rightarrow\ \text{x}=(\text{y}-4)^\frac{1}{3}$
$\therefore\ \text{f}^{-1}(\text{x})=(\text{x}-4)^\frac{1}{3}$
$\therefore\ \text{f}^{-1}(3)=(3-4)^\frac{1}{3}=-1$

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