How do you find $x$ such as $y \leq \sqrt{2x-x^2}$?

Question

I am unable to isolate the variable $x$ of this inequality $y \leq \sqrt{2x-x^2}$ ( where $0 \leq y \leq 1 $)

Is it correct doing this: $y^2 \leq 2x-x^2$? I found that $y^2 \leq x \leq 2-y^2$ and $0 \leq x \leq 2$. Is it correct?

From here I am not sure how to proceed. Thanks in advance for any help.

Answer 1

Let $f(x)=\sqrt{2x-x^2}$

its domain is $[0,2]$.

$$f'(x)=\frac{1-x}{f(x)}$$

the maximum is $f(1)=1$.

Let $y\in[0,1]$

we look for $x$ such that $$x^2-2x+y^2=0$$

which gives two solutions, one in $[0,1]$ and the other in $[1,2]$ : $$x=1\pm\sqrt{1-y^2}$$

Answer 2

When you get to $x^2-2x+y^2$ = 0, you can complete the square for the x-terms and get $$ (x^2-2x+1) + y^2 = 1$$ $$ (x-1)^2 + y^2 = 1$$

This is a circle of radius 1 centered at (1,0). So now you just need to test 2 regions, the inside of the circle and outside of the circle. Thus, take a point in each region and see if the inequality holds. Try the center of the circle and a point on one of the axes that lies outside the circle.

Answer 3

First, you have to determine the domain of validity of this inequation: $$x^2-2x\ge 0\iff x(2-x)\ge 0\iff x\in[0,2].$$ Next

• if $y\le 0$, the inequation is satisfied for all $x\in[0,2]$,
• if $y>0$, the inequation is equivalent (on $[0,2]$) to $y^2\le 2x-x^2$, i.e. to $$p_y(x)=x^2-2x+y^2\le 0.$$ As a quadratic inequation in $x$, it is satisfied if and only if it has real roots, $\xi_0\le\xi_1$ and $\;x\in[0,2]\cap[\xi_0,\xi_1]$. It has real roots if and only if its reduced discriminant $\;\Delta'=1-y^2\ge 0$, i.e. $0<y\le1$ in the present case. So we need to place the numbers $\;0,2,\xi_0=1-\sqrt{1-y^2},\xi_1=1+\sqrt{1-y^2},2\;$ w.r.t. each other.

Now $p_y(0)=p_y(2)=y^2\ge 0$, so neither $0$ nor $2$ separates the roots. Furthermore,by Vieta's relations, $\dfrac{\xi_0+\xi_1}2=1$, so that $\;0\le\xi_0 <1<\xi_1 \le 2\;$, and finally the solutions are $$[\xi_0,\xi_1]\quad \text{if }\;0<y\le 1.$$

Answer 4

Since you're assuming $y\ge0$, the given inequality is equivalent to $y^2\le2x-x^2$, that is, $x^2-2x+y^2\le0$.

The discriminant of $x^2-2x+y^2$ is $4(1-y^2)$. If $y>1$, the inequality $x^2-2x+y^2\le0$ is satisfied for no $x$.

If $y\le1$, the inequality is satisfied for $1-\sqrt{1-y^2}\le x\le 1+\sqrt{1-y^2}$. However we still need $x\ge y$; note that $1-\sqrt{1-y^2}\ge y$ (prove it). Thus the solutions are $$ y\le x\le 1+\sqrt{1-y^2}\qquad (\text{for $y\le1$}) $$

The following graph confirms this.