2

Let $M$ be the subset of $\mathbb{R}^2$ bounded by the parabolas $y = 2x^2$ and $y = x^2 + 1$. Express $M$ as a general region w.r.t $y$-axis and w.r.t $x$-axis.

I tried using the definition for $M$ w.r.t $y$-axis: $$M = \{ (x,y) \in \mathbb{R}^2, \, a \le x \le b, \, f_1(x) \le y \le f_2(x) \}$$ where $a, b \in \mathbb{R}$ anf $f_1, f_2$ - continuous.

However, for my problem it seems that $x$ cannot be bounded between two real numbers.

Thank you in advance!

George R.
  • 2,833

1 Answers1

0

$a$ and $b$ (with $a\leq b$) are the $x$-values of where the functions intersect, so the solutions of $f_1(x)=f_2(x)$, so solving $2x^2=x^2+1$ gives us $a=-1$ and $b=1$.

The Phenotype
  • 5,199
  • What about the general region w.r.t $x$-axis? – George R. Jan 01 '18 at 17:15
  • Do you mean when $f_1$ and $f_2$ don't intersect (or only once)? In that case you just throw away the $a$- or $b$-inequality. – The Phenotype Jan 01 '18 at 17:19
  • I mean to find $M$ but this time put $y$ between two real constants and bound $x$ between two functions in $y$. – George R. Jan 04 '18 at 19:18
  • Then you find the (zero, one or) two corresponding solutions of $f_1(y)=f_2(y)$ to obtain two solutions $y_1$ and $y_2$, now these are the bounds for $y$. It is practically the same (due to symmetry in $x=y$). – The Phenotype Jan 04 '18 at 19:25