functional-inequalitiesinequality
$\sqrt{3x-x^2}<4-x$
I know I can't simply square both sides of an inequality.
I have narrowed down the possible values of x =>
x belongs to [0,3] because the expression inside the square root cannot be negative. (Ignoring imaginary numbers)
And strangely enough, that is the given solution! => x belongs to [0,3]
Is the solution correct? And how? Doesn't the RHS matter?
Let $x\ge 0$. Then, by definition, $\sqrt{x}$ is the non-negative number whose square is $x$.
The function $f(x)=\sqrt{x}$ is an increasing function. Thus, if $p$ and $q$ are non-negative, then $p\lt q$ iff $\sqrt{p}\lt \sqrt{q}$.
Remark: Regrettably, it is not uncommon in the schools for teachers, and texts, to write, for example, $\sqrt{9}=\pm 3$. The convention when functions are studied is that $\sqrt{x}$ is non-negative. This is done so that $\sqrt{x}$ will be a function of $x$. Recall that for $f$ to be a function, we cannot have two different values of $f$ at a number $a$.
To Henry's comment here is the drawing:
Best Answer
As you said, $x \in [0,3]$. Now if $x \in [0,3]$ the square root is well defined, and hence we have $0 \leq \sqrt {3x-x^{2}}< 4-x$. (You can square both sides since the square function is strictly increasing on $[0, \infty]$ Now squaring both sides we get $3x-x^{2}< x^{2}-8x+16$ hence $2x^{2}-11x+16>0$ but the discriminant of $2x^{2}-11x+16$ is strictly negative, and since the leading term $2>0$ then $2x^{2}-11x+16>0$ for all $x \in [0,3]$. Hence $[0,3]$ is the solution set.