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

Exercise

Determine the domain of the function:

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

Final Answer

Show 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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