I got a question given by my lecturer to find the domain and range of this function: $$y=\sqrt{2-x^2}$$
For the domain, we start by restricting the square root to be non-negative. After factorising, it becomes $(x+\sqrt2)(x-\sqrt2)\le0$. This is my explanation.
But the lecturer at this point jumps to the conclusion that $y:[-\sqrt2,\sqrt2]\mapsto[0,2]$ without any explanation.
I tried to substitute 2 as the value of y but it is wrong. Can anyone care to explain how he derived the domain from the factorisation and range?
$\endgroup$3 Answers
$\begingroup$The correct factorization is
$2-x^2=(\sqrt{2}+x)(\sqrt{2}-x)$
where the left factor is nonnegative for $x\geq-\sqrt{2}$ and the right factor is nonnegative for $x\leq\sqrt{2}$. Thus the function is defined on $[-\sqrt{2},\sqrt{2}]$. The factors can´t be both negative since the conditions $x<-\sqrt{2}$ and $\sqrt{2}<x$ exclude each other. Thus $[-\sqrt{2},\sqrt{2}]$ is the maximal set of real numbers where the function is defined. Obviously the minimal value $0$ is taken for $x\in \{-\sqrt{2},\sqrt{2}\}$ and because of the monotonicity of the squareroot the maximal value is taken when the radicand is maximal thus for $x=0$ and this maximal value is $\sqrt{2}$. Thus the mapping is a mapping $[-\sqrt{2},\sqrt{2}]\rightarrow [0,\sqrt{2}]$ The image of $f$ is $Im(f)=[0,\sqrt{2}]$.
$\endgroup$ 3 $\begingroup$Your lecturer appears to have made a typo.
it should have been $[-\sqrt{2},\sqrt{2}] \rightarrow [0,\sqrt{2}]$
You have made the correct start by saying that the square root must be non-negative. Using that information you can then change your equation from this: $$y = \sqrt{2-x^2}$$ to this: $$2-x^2 \geq0$$
a bit of rearranging and out pops the domain limits. $$2 \geq x^2$$
$$-\sqrt{2} \leq x \leq \sqrt{2}$$
so from this we can see the domain limits are $[-\sqrt{2}, \sqrt{2}]$
Then to get the upper bound on the range, maximise the radicand (the part under the square root). $x=0 \rightarrow y=\sqrt{2}$
So you end up with $[-\sqrt{2},\sqrt{2}] \rightarrow [0,\sqrt{2}]$
$\endgroup$ 4 $\begingroup$Anyway, it's much simpler to use absolute values: $$2-x^2\ge 0\iff x^2\le 2\iff \sqrt{x^2}=\lvert x\rvert\le\sqrt 2\iff -\sqrt 2\le x\le\sqrt2.$$
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