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

Exercise

Determine the domain of the function:

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

$$x\geq 2\text{ or } x\leq\frac{1}{2}$$

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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