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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


-1<x\leq 1\text{  or  } 2\leq x< 3

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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