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A function is one-one if $f(x_1) = f(x_2)$ implies $x_1 = x_2$. Let us check for $f(x) = x^2 - 1$. Consider $x_1 = 1$ and $x_2 = -1$. Both are in the domain $R$. $$f(1) = (1)^2 - 1 = 0$$ $$f(-1) = (-1)^2 - 1 = 0$$ Since $f(1) = f(-1)$ but $1 \neq -1$, the function is not one-one.
A function is onto if for every $y$ in the codomain $R$, there exists an $x$ in the domain $R$ such that $f(x) = y$. Let $y = -2$. $$x^2 - 1 = -2$$ $$x^2 = -1$$ Since $x^2 = -1$ has no real solution, there is no $x \in R$ such that $f(x) = -2$. Thus, the function is not onto.
Final Answer: The function is neither one-one nor onto.
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