Domain of One Variable Function – A function with log – Exercise 5749


Determine the domain of the function:

y=\sqrt{\log_3 [(3-2x)(1-x)]}

Final Answer

x\geq 2\text{  or  } x\leq\frac{1}{2}


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:


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:


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


Because it is broken down into factors, its roots are easy to find. The first root is




The second root is



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


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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