The function f: mathbb{R} to mathbb{R} define | vedclass

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(D) Given the function $f(x) = x^2 + x$. Step $1$: Check for one-one or many-one. We observe that $f(0) = 0^2 + 0 = 0$ and $f(-1) = (-1)^2 + (-1) = 1

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Many-one into

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(D) Given the function $f(x) = x^2 + x$. Step $1$: Check for one-one or many-one. We observe that $f(0) = 0^2 + 0 = 0$ and $f(-1) = (-1)^2 + (-1) = 1 - 1 = 0$. Since $f(0) = f(-1)$ but $0 
eq -1$,the function is many-one. Step $2$: Check for onto or into. We can rewrite the function by completing the square: $f(x) = x^2 + x + \frac{1}{4} - \frac{1}{4} = (x + \frac{1}{2})^2 - \frac{1}{4}$. Since $(x + \frac{1}{2})^2 \ge 0$ for all $x \in \mathbb{R}$,it follows that $f(x) \ge -\frac{1}{4}$. This means the range of $f$ is $[-\frac{1}{4}, \infty)$. Since the range $[-\frac{1}{4}, \infty)$ is not equal to the codomain $\mathbb{R}$,the function is into. Conclusion: The function is many-one into.

The function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^2 + x$ is:

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