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To prove that the function $f(x) = 4x^3 - 5$ is one-one, we need to show that if $f(x_1) = f(x_2)$, then $x_1 = x_2$ for all $x_1, x_2 \in R$.
Let $f(x_1) = f(x_2)$. Then, $$4x_1^3 - 5 = 4x_2^3 - 5$$ $$4x_1^3 = 4x_2^3$$ $$x_1^3 = x_2^3$$ Taking the cube root of both sides, we get $$x_1 = x_2$$ Thus, $f(x)$ is one-one.
To prove that the function $f(x) = 4x^3 - 5$ is onto, we need to show that for every $y \in R$, there exists an $x \in R$ such that $f(x) = y$.
Let $y = f(x)$. Then, $$y = 4x^3 - 5$$ $$y + 5 = 4x^3$$ $$x^3 = \frac{y + 5}{4}$$ $$x = \sqrt[3]{\frac{y + 5}{4}}$$ Since $y \in R$, $\frac{y + 5}{4} \in R$, and the cube root of any real number is a real number, $x \in R$.
Now, we verify that $f(x) = y$: $$f(x) = f\left(\sqrt[3]{\frac{y + 5}{4}}\right) = 4\left(\sqrt[3]{\frac{y + 5}{4}}\right)^3 - 5 = 4\left(\frac{y + 5}{4}\right) - 5 = y + 5 - 5 = y$$ Thus, for every $y \in R$, there exists an $x \in R$ such that $f(x) = y$. Therefore, $f(x)$ is onto.
Since $f(x)$ is both one-one and onto, it is a bijective function.
Final Answer: The function $f(x) = 4x^3 - 5$ is one-one and onto.
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