Analysis for each problem on Quiz (2):

Analysis for Question 1:

Please answer the following questions based on the implicitly defined function $|x-y^2|=1-|x|$:

Which aspect should we consider when finding the domain of the given function?

$|x|$

$|x-y^2|$

We need calculus help

We need to solve the equation $|x-y^2|=1-|x|$

$y^2 $

In any question that wants us to determine the domain of any given function that absolute value function(s), or square terms involved, these are the terms that help us to figure out the domain for the given function, so the correct answers should include the three of the below:

[MATH]\color{yellow}\bbox[5px,purple]{|x|}[/MATH], [MATH]\color{yellow}\bbox[5px,grey]{|x-y^2|}[/MATH] and [MATH]\color{yellow}\bbox[5px,blue]{y^2 }[/MATH].

In fact, we can draw a quick and useful partial conclusion since we know the absolute value of any number is always greater than or equal to 0.

So $|x-y^2|\ge 0\,\implies\,\,1-|x|\ge 0$ This further implies [MATH]\color{green}\bbox[5px,yellow]{-1\le x \le 1}[/MATH].

Analysis for Question 2:

Question 2: Which of the following definitions is true?

$A. 1-x=\begin{cases}x-y^2, & x-y^2\ge 0 \\ -(x-y^2), & x-y^2<0 \\ \end{cases}$ or

$B. 1-|x|=\begin{cases}x-y^2, & x-y^2\ge 0 \\ -(x-y^2), & x-y^2<0 \\ \end{cases}$

We are all very familiar about how to remove the absolute value bars on function like $y=|x|$ by rewriting the function of $y$ as a piecewise function:

$y=|x|=\begin{cases}x, & x\ge 0 \\ -(x), & x<0 \\ \end{cases}$

So, we do the same with the function that we're given, i.e.$|x-y^2|=1-|x|$, it might not be immediately obvious for you how you are going to relate the above $y=|x|$ with $|x-y^2|=1-|x|$.

That is fine, and we are not stuck yet, what we are going to do next is to further examine the piecewise function that we obtained from $y=|x|$:

Note that as long as we have isolated the any one of the absolute value function on either side, then we can began to redefine our function. Yes, we notice that we have now two absolute value functions, ( [MATH]\color{yellow}\bbox[5px,purple]{|x|}[/MATH] and [MATH]\color{yellow}\bbox[5px,grey]{|x-y²|}[/MATH]). Therefore you don't know for certain if you should work with:

$1-|x|=|x-y^2|$, and end up as

$1-|x|=\begin{cases}x-y^2, & x-y^2\ge 0 \\ -(x-y^2), & x-y^2<0 \\ \end{cases}$

or

But, if you start with $1-|x-y^2|=|x|$, you deal with:

$1-|x-y^2|=\begin{cases}x, & x\ge 0 \\ -(x), & x<0 \\ \end{cases}$

If you work with $1-|x|=|x-y^2|$, then we have:

$1-|x|=\begin{cases}x-y^2, & x-y^2\ge 0 \\ -(x-y^2), & x-y^2<0 \\ \end{cases}$

But, if you start with $1-|x-y^2|=|x|$, you deal with:

$1-|x-y^2|=\begin{cases}x, & x\ge 0 \\ -(x), & x<0 \\ \end{cases}$

Our quiz(2) problems revolve around the first option so we will work with that instead. So the correct answer for number 2 is B.

It is worth noticing that the first choice has the left side expression of $1-x$, but we could never remove the absolute value bars without doing anything to it, so, the first choice is actually out of the question and it cannot be the answer and one can simply rule that out. :D

Analysis for Question 3: Considering the part where $x-y^2<0$, what can we further conclude for the interval(s) of the domain of this given function?

A. $-1\le x \le 0$

B. $-1\le x \le -\dfrac{1}{2}$

C. $-\dfrac{1}{2}\le x <0$ and $0\le x \le -\dfrac{1}{2}$

D. $-\dfrac{1}{2}\le x \le 0$

From what we just defined, we have

$1-|x|=\begin{cases}x-y^2, & x-y^2\ge 0 \\ -(x-y^2), & x-y^2<0 \\ \end{cases}$

The question asked us to consider the interval where $x-y^2<0$, so here is the function we need to examine:

$1-|x|=-(x-y^2)$

We need to remember the first observation that we obtained, which is [MATH]\color{green}\bbox[5px,yellow]{-1\le x \le 1}[/MATH].

Now, to remove the absolute value bars around $x$, we have two cases to think here:

$-(x-y^2)=\begin{cases}1-x, & x\ge 0 \,\,\,\,\text{and}\,\,-1\le x \le 1\implies\,\,0\le x \le 1\\ 1-(-x), & x<0 \,\,\,\,\text{and}\,\,-1\le x \le 1\implies\,\,-1\le x <0\\ \end{cases}$

For the first case, we see that we have

$-(x-y^2)=1-x$

$-x+y^2=1-x$

$y^2=1$

$y=\pm 1$ on the interval $0\le x \le 1$.

For the second case, observe that

$-(x-y^2)=1-(-x)$

$-x+y^2=1+x$

$y^2=2x+1$

But $y^2\ge 0$, so $2x+1\ge 0\,\implies\,x\ge -\dfrac{1}{2}$ and now the interval that applied to the existing case becomes $-\dfrac{1}{2}\le x<0$ for the function $y^2=2x+1$.

By combining the results of two sub-intervals we found by considering $x-y^2<0$, the answer for question 3 is C.

Analysis for Question 4: Repeat question 3 but now considering the part where $x-y^2>0$.

A. $0\le x \le 1$

B. $\dfrac{1}{2}\le x \le 1$

C. $-\dfrac{1}{2}\le x <1$

D. $-1\le x \le 1$

From what we just defined, we have

$1-|x|=\begin{cases}x-y^2, & x-y^2\ge 0 \\ -(x-y^2), & x-y^2<0 \\ \end{cases}$

The question asked us to consider the interval where $x-y^2\ge 0$, so here is the function we need to examine:

$1-|x|=x-y^2$

We also need to remember the first observation that we obtained, which is [MATH]\color{green}\bbox[5px,yellow]{-1\le x \le 1}[/MATH].

By the similar token, to remove the absolute value bars around $x$, we have two cases to think here:

$x-y^2=\begin{cases}1-x, & x\ge 0 \,\,\,\,\text{and}\,\,-1\le x \le 1\implies\,\,0\le x \le 1\\ 1-(-x), & x<0 \,\,\,\,\text{and}\,\,-1\le x \le 1\implies\,\,-1\le x <0\\ \end{cases}$

For the first case, we see that we have

$x-y^2=1-x$

$y^2=2x-1$

But $y^2\ge 0$ therefore $2x-1\ge 0\implies x\ge \dfrac{1}{2}$ and the interval that applied to this case is hence $\dfrac{1}{2}\le x\le 1$.

For the second case, observe that

$x-y^2=1-(-x)$

$x-y^2=1+x$

$y^2=-1$

Therefore, we can say there is nothing valid from this case that is worth considering for.

The answer for this question 4 is hence B, $\dfrac{1}{2}\le x\le 1$.

Analysis for Question 5: Which of the following diagram represents the curve $|x-y^2|=1-|x|$?

By doing a summary for what we have collected so far, we see we have broken down the given original function into three intervals, each of which has different function that governs it:

For $-\dfrac{1}{2}\le x <0$, we have $y^2=2x+1$.

For $0\le x\le 1$, we have $y=-1$ and $y=1$.

For $\dfrac{1}{2}\le x\le 1$, we have $y^2=2x-1$.

By putting them altogether on the same coordinate plane, it is not hard to see that B is the diagram represents the curve $|x-y^2|=1-|x|$.

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