 # Domain of One Variable Function – A function with polynomials inside square roots – Exercise 5752

Exercise

Determine the domain of the function

$$y=\sqrt{x^2-3x+2}+\frac{1}{\sqrt{3+2x-x^2}}$$

$$-1

Solution

Given the function:

$$y=\sqrt{x^2-3x+2}+\frac{1}{\sqrt{3+2x-x^2}}$$

We find the domain of the function. Because there is a denominator, the expression in the denominator must be different from zero:

$$\sqrt{3+2x-x^2}\neq 0$$

Also, there are square roots, so the expressions inside the roots must be non-negative:

$$3+2x-x^2\geq 0$$

$$x^2-3x+2\geq 0$$

We solve the last inequality:

$$x^2-3x+2\geq 0$$

It’s a square inequality. Let’s look at the square equation:

$$x^2-3x+2=0$$

Its coefficients are

$$a=1, b=-3, c=2$$

The coefficient of the squared expression (a) is positive, so the parabola (quadratic equation graph) “smiles” (= bowl-shaped). The sign of the inequality means we are looking for the sections the parabola is above the x-axis. We find the solutions (= zeros = roots) of the quadratic equation using the quadratic formula. Putting the coefficients in the formula gives us

$$x_{1,2}=\frac{3\pm \sqrt{{(-3)}^2-4\cdot 1\cdot 2}}{2\cdot 1}=$$

$$=\frac{3\pm \sqrt{1}}{2}=$$

$$=\frac{3\pm 1}{2}$$

Hence, we get the solutions:

$$x_1=\frac{3+ 1}{2}=2$$

$$x_2=\frac{3-1}{2}=1$$

Because we are looking for the section above the x-axis or on it and the parabola “smiles”, we get that the solution of the inequality is

$$x\leq 1\text{ or } x\geq 2$$

The other two inequalities:

$$\sqrt{3+2x-x^2}\neq 0$$

$$3+2x-x^2\geq 0$$

are equivalent to the inequality:

$$3+2x-x^2>0$$

It is a square inequality. Let’s look at the quadratic equation:

$$-x^2+2x+3=0$$

Its coefficients are

$$a=-1, b=2, c=3$$

The coefficient of the squared expression (a) is negative, so the parabola (quadratic equation graph) “cries” (= bowl-shaped). The sign of the inequality means we are looking for the sections the parabola is above the x-axis. We find the solutions (= zeros = roots) of the quadratic equation using the quadratic formula. Putting the coefficients in the formula gives us

$$x_{1,2}=\frac{-2\pm \sqrt{2^2-4\cdot (-1)\cdot 3}}{2\cdot (-1)}=$$

$$=\frac{-2\pm \sqrt{16}}{-2}=$$

$$=\frac{-2\pm 4}{-2}$$

Hence, we get the solutions:

$$x_1=\frac{-2-4}{-2}=3$$

$$x_2=\frac{-2+4}{-2}=-1$$

Because we are looking for the section above the x-axis and the parabola “cries”, we get that the solution of the inequality is

$$-1

Intersect the results, meaning

$$x\leq 1\text{ or } x\geq 2$$

and

$$-1

$$-1