algebra-precalculus
$f(x) = -1 + \sqrt{x+2}$
$$Dom(f(x)) = \{x|x\geq-2\}$$
$$Ran(f(x)) = \{y|y\geq-1\}$$
To see that the range is, all of the set above, it needs to be shown that for $y\geq-1$ there exists a real number x such that,
$$y = -1 +\sqrt{x+2}$$
Solving for x we have,
$$y+1 = \sqrt{x+2} \\
x+2 = (y+1)^2 \\
x = (y+1)^2 – 2 $$
this shows that the domain is the all the set $\{y|y\geq -1\}$
Is this correct?
Also, is any of my notation incorrect or can it be improved?
Thanks
You are wrong about the domain of $f$.
$f(x)=x^2$ has the domain $(-\infty,\infty)$. Therefore $Ran(g)$ is actually a subset of $Dom(f)$.
The domains of $f(g(x))$ and $g(f(x))$ are both $(-\infty,\infty)$: i.e. $\mathbb R$, all real numbers.
The range of $f(g(x))=(\sin(x))^2$ is $[0,1]$. The zero is attained at $x=0$ and the one at $x=\frac{\pi}2$. The continuity of the functions guarantees that all intermediate values are also attained.
The range of $g(f(x))=\sin(x^2)$ is $[-1,1]$. The minus-one is attained at $x=\sqrt{\frac{3\pi}2}$ and the one at $x=\sqrt{\frac\pi 2}$. The continuity of the functions guarantees that all intermediate values are also attained.
I am going to assume (you should really specify this) that $x,y\in\mathbb{R}$. The domain is rather straightforward since all you need is the expression within the root sign to be non-negative. That is, you need $1+x-y^2\geq0$ or "in more simple terms" $1+x\geq y^2$. So $$ \operatorname{Dom}\bigl(f(x,y)\bigr)=\{1+x\geq y^2\mid x,y\in\mathbb{R}\}. $$ The range is slightly more tricky but not so much. Since $x$ and $y$ are allowed to be anything in $\mathbb{R}$ (so long as they are in the domain), we can work out a way for all of non-negative $\mathbb{R}$ to be mapped to. That is, $$ \operatorname{Rng}\bigl(f(x,y)\bigr)=\mathbb{R}_{\geq0}. $$ For example, how could you get $f(x,y)=e$ for some values of $x$ and $y$? What if $x=y^2-1+e^2$? Then we have that $f(x,y)=\sqrt{1+x-y^2}=\sqrt{1+(y^2-1+e^2)-y^2}=\sqrt{e^2}=e$.
Can you figure it out from here?
Best Answer
One step missing in the end is to check that $f((y+1)^2-2) = y$ when $y\geq -1$, as Siong Thye Goh points out. This is important, since if $y<-1$, $f((y+1)^2-2) = -y-2$, actually!
To be more clear what I mean, you basically wrote: "if $y = -1 + \sqrt{x+2}$ has a solution for some $y$, then that solution is $x = (y+1)^2-2$", but you didn't comment that in the case $y<-1$ there is no solution and you didn't check that in the case $y\geq -1$ there really exists a solution (as I wrote in the 1st paragraph).
Thus, your phrasing is not as clear as it could be.
Here's one alternative approach to be more precise. As soon as you wrote $\sqrt{\,\cdot\,}$, it is implicit that by this you mean a bijective function $[0,\infty)\to [0,\infty)$, inverse to $x\mapsto x^2\colon [0,\infty)\to [0,\infty)$.
Let's define $g(x) = -1 + x$, $h(x) = \sqrt x$ and $k(x) = x + 2$. Then, $f = g\circ h\circ k$ and the range of $f$ is
$$f([-2,\infty)) = g(h(k([-2,\infty)))= g(h([0,\infty))) = g([0,\infty)) = [-1,\infty).$$