Exercise
Determine the domain of the function:
y=\sqrt{\log_3 [(3-2x)(1-x)]}
Final Answer
Solution
Let’s find the domain of the function:
y=\sqrt{\log_3 [(3-2x)(1-x)]}
Because there is a log, we need the expression inside the log to be greater than zero:
(3-2x)(1-x)>0
It also has a square root, so the expression inside the root must be non-negative:
\log_3 [(3-2x)(1-x)]\geq 0
Solve the first inequality:
(3-2x)(1-x)>0
It is a square inequality. Let’s look at the quadratic equation:
(3-2x)(1-x)=0
Because it is broken down into factors, its roots are easy to find. The first root is
3-2x=0
3=2x
x=\frac{3}{2}
The second root is
1-x=0
x=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>\frac{3}{2}\text{ or } x<1
Solve the second inequality:
\log_3 [(3-2x)(1-x)]\geq 0
We got a log in the inequality. By log definition, we get that our inequality is equivalent to:
(3-2x)(1-x)\geq 3^0=1
Note: If the log base was less than one, we would turn over the inequality sign.
Open brackets:
3-3x-2x+2x^2\geq 1
2x^2-5x+3\geq 1
2x^2-5x+2\geq 0
It is a square inequality. Its coefficients are
a=2, b=-5, 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{5\pm \sqrt{{(-5)}^2-4\cdot 2\cdot 2}}{2\cdot 2}=
=\frac{5\pm \sqrt{9}}{4}=
=\frac{5\pm 3}{4}
Hence, we get the solutions:
x_1=\frac{5+ 3}{4}=2
x_2=\frac{5- 3}{4}=\frac{1}{2}
Since the parabola “smiles” and we are interested in the sections above the x-axis or on it, we get
x\geq 2\text{ or } x\leq\frac{1}{2}
Finally, we intersect both results ( “and”) and get:
x>\frac{3}{2}\text{ or } x<1
and
x\geq 2\text{ or } x\leq\frac{1}{2}
The intersection is the final answer:
x\geq 2\text{ or } x\leq\frac{1}{2}
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