I'm having trouble with doing two bijective proofs. I understand bijection and how it works, but I'm just unsure how to word the proof using formulas to find specific values were function are/aren't 1-1/onto.
I have two functions: $f(x) = -3x^2+7$ and $f(x) = \dfrac{x+1}{x+2}$.
I know the first one is not bijective since the $y$-value range only goes from $-\infty$ to $7$ (not onto) and being a parabola, all $y$-values except $7$ can be achieved with two different $x$-values (not $1$-$1$).
The second one also isn't bijective since the function is undefined at $x=-2$ (not $1$-$1$) and isn't onto being undefined at $y=1$ (not onto).
Best Answer
What you’ve written is reasonably clear, but it could certainly be tidied up.
$x^2\ge 0$ for all $x\in\Bbb R$, so $-3x^2\le 0$, and $f(x)=-3x^2+7\le 7$ for all $x\in\Bbb R$. Clearly, then, $8$ is not in the range of $f$, and $f$ is not onto. Moreover, $f(1)=4=f(-1)$, so $f$ is not $1$-to-$1$. (You could of course use different specific examples; I just picked very handy ones.)
I have to take back part of what I said in my comment. Let $$f(x)=\frac{x+1}{x+2}=\frac{(x+2)-1}{x+2}=1-\frac1{x+2}\;.$$ Clearly $f(x)$ is defined for all real $x$ except $-2$. It’s also clear that if $x\ne-2$, then $\frac1{x+2}\ne 0$ and hence $f(x)\ne 1$, so $1$ is not in the range of $f$. Thus, $f$ is not a bijection from $\Bbb R$ to $\Bbb R$, since neither its domain nor its range is all of $\Bbb R$. However, $f$ is a bijection from $\Bbb R\setminus\{-2\}$ to $\Bbb R\setminus\{1\}$. We already know that $f$ is defined on $\Bbb R\setminus\{-2\}$. Suppose that $y\in\Bbb R\setminus\{1\}$; then $y$ is in the range of $f$ if and only if the equation $y=1-\frac1{x+2}$ has a solution, which it has: it’s equivalent to $\frac1{x+2}=1-y$ and thence to $x+2=\frac1{1-y}$ and $x=\frac1{1-y}-2$, which is indeed defined, since $y\ne 1$. This calculation shows not only that $\Bbb R\setminus\{1\}$ is the range of $f$ but also that $f$, considered as a function from $\Bbb R\setminus\{-2\}$ to $\Bbb R\setminus\{1\}$, has an inverse, $$f^{-1}(x)=\frac1{1-x}-2\;,$$ and is therefore a bijection. Finally, its restriction to any subset of $\Bbb R$ on which it’s defined is $1$-to-$1$.