Question

a) Which of the following functions are 1-1 and which are onto? Justify your answer.
i) f : R!R0 given by f (x) = x2 where R0 is the set fx 2 Rjx 0g.
ii) f : R!R given by f (x) = x2+x+1.

Answer 1

Answer on Question #40792– Math - Linear Algebra

a) Which of the following functions are 1-1 and which are onto? Justify your answer.

i) $f: R \to R_0$ given by $f(x) = x^2$ where $R_0$ is the set for $f(x)$ 2 Rjx 0g.

ii) $f: R \to R$ given by $f(x) = x^2 + x + 1$.

Solution:

The function is injective or 1 to 1 if every element of the function's codomain is the image of at most one element of its domain.

The function $f$ from a set $X$ to a set $Y$ is surjective (or onto), or a surjection, if every element $y$ in $Y$ has a corresponding element $x$ in $X$ such that $f(x) = y$.

i) $f: R \to R_0$ given by $f(x) = x^2$

If $R_0 = R \setminus \{0\}$ then $f$ is not onto because for $y = -2$ we can't find a $x$ such that $f(x) = x^2$ and it is not 1 to 1, because $f(1) = f(-1) = 1$.

If $R_0 = [0; +\infty)$ then $f$ is onto, because for every $y$ in $[0; +\infty)$ exists $x = \sqrt{y}$ such that $y = f(x)$ and it is not 1 to 1, because $f(-1) = f(1)$.

ii) $f: R \to R$ given by $f(x) = x^2 + x + 1$

This function is not onto, because for $y = 0$ we can't find $x$ such that $0 = x^2 + x + 1$.

And it is not 1 to 1 because $f(1) = f(-2) = 3$.