Find the domain and the range of a function. I am puzzled. 
My objective is to find the domain and the range of the following function:
$$ f(x) ={\sqrt{3x -x^{2}-2}} .$$
I can immediately see that the expression in the square root  of cannot be less than $0$. Then I proceed by finding where the expression $$ f(x) ={3x -x^{2}-2} $$ is $$0 =3x -x^{2}-2$$ with the PQ-formula.
I later get the following values for x when the function is equal to zero.
$$ x_1 = 1 , x_2=2$$
I proceed to make a sign diagram in order for me to see where the expression is positive or zero, since, as stated earlier, that is where the domain of the function will be:
$$ x    \in  [1,2].$$
But at this point, I have no clue as to how to find the range. I mean sure, I know that the lower boundary is zero, but the maximum? I can only think of solving this problem by finding the derivative and a critical point. But this course is fairly easy, and I am supposed to use a less advanced method, and I think it has to do with the symmetry of the expression. But I have no clue on how what to do...
 A: Hint:
$f(x)=\sqrt{\left(\dfrac12\right)^2-\left(x-\dfrac32\right)^2}$
A: Hint:
The quadratic polynomial $-x^2+3x-2$ is positive between its roots since its leading coefficient is negative. Furthermore, it is known from high school that a quadratic polynomial $ax^2+bx+c$ has a single extremum, attained at $x=-\dfrac{b}{2a}$. Here this means the extremum is attained at $\frac32$, and it is
$$-\frac 94+\frac 92-2=\frac 14.$$
Thus, the range is $\Bigl[0,\frac12\Bigr]$.
A: The equation $-x^2+3x-2$ is just a parabola with the concavity donward, so the polynomial is positive when $1\leq x\leq 2$: this is the domain. Now after a few steps you can write your equation in the form: $$y=\sqrt{\left(\dfrac12\right)^2-\left(x-\dfrac32\right)^2}$$ This is clearly an arc of circle so the range is $y\geq 0$ and more precisely $0\leq y\leq\frac{1}{2}$ because it's a semicicle with centre on the $x$-axis.
A: $3x- x^2- 2= 0$ is the same as $3x- x^2= -(x^2- 3x+ 9/4- 9/4)= 9/4- (x+ 3/2)$.  That has its maximum value, 9/4, when x= -3/2.  The maximum value of the square root is 3/2 so the range is from 0 to 3/2.
