Exercise
Determine the domain of the function
y=\sqrt{x^2-3x+2}+\frac{1}{\sqrt{3+2x-x^2}}
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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