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

Exercise

Determine the domain of the function

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

Final Answer

Show final answer

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

$-1<x<3$

Intersect the results, meaning

$x\leq 1\text{  or  } x\geq 2$

and

$-1<x<3$

lead to the final answer:

$-1<x\leq 1\text{  or  } 2\leq x< 3$

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