Finding domain of implicit functions.
Suppose $f(x,y)=k$ where k is a constant. I am supposed to find the domain of $ f(x,y)=1$ so should I check for the value of $x $where $f(x,y)$ is defined or first write $ y=f(x) $and then find the values of x for which f(x) is defined. Consider $x^{2y}=1$. The domain will be all real x I think but the answer in my book is given as only $x>0 $or $ x=0.$
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$\begingroup$Take $x=-1,\;y=1/2.$ Then $x^{2y}=-1$ and not $1.$ Given any negative $x$-value you can choose a $y$ which makes the equality fail.
$\endgroup$ 1 $\begingroup$The given function $$f(x,y):=x^{2y}$$ has as domain the right half plane $$H:=\bigl\{(x,y)\in{\mathbb R}^2\bigm| x>0, \ -\infty<y<\infty\bigr\}\ ,$$ because as soon as you consider powers $b^t$ with variable noninteger exponents the base $b$ is automatically restricted to ${\mathbb R}_{>0}$.
Let's see whether the equation $f(x,y)=1$, i.e., $x^{2y}=1$, "defines $y$ as a function of $x$" in some way. Note that by definition $$1=x^{2y}:=\exp(2y\log x)\ ,$$ and as $\exp$ is monotonically increasing this is equivalent with $2y\log x=0$. It follows that $$y=0\quad\vee\quad \log x=0,\quad{\rm resp.,}\quad y=0\quad\vee\quad x=1\ .$$ In other words the solution set $$\bigl\{(x,y)\in H\bigm| f(x,y)=1\bigr\}$$ consists of the horizontal half line $y=0$ together with the vertical line $x=1$.
We therefore can say the following: The given equation defines implicitly the function $y=\phi(x):\equiv0$ separately in the $x$-intervals $(0,1)$ and $(1,\infty)$.
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