Domain of One Variable Function – A function with square root and ln – Exercise 5736

Exercise

Determine the domain of the function:

$y=\sqrt{x^2-1}+\ln\sqrt{1-4x^2}$

Final Answer

Show final answer

Solution

Let’s find the domain of the function:

$y=\sqrt{x^2-1}+\ln\sqrt{1-4x^2}$

Because there are square roots, the expressions inside the roots must be non-negative:

$x^2-1\geq 0$

$1-4x^2\geq 0$

Also, there is an ln function, so we need the expression inside the ln to be positive:

$\sqrt{1-4x^2}>0$

We got 3 inequalities. Let’s solve the first inequality:

$x^2-1\geq 0$

It is a square inequality. The roots of the quadratic equation:

$x^2-1=0$

are

$x=\pm 1$

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

$-1\geq x \text{  or  } x\geq 1$

The other two inequalities are equivalent to the inequality:

$1-4x^2>0$

Again, it is a square inequality. The roots of the quadratic equation:

$1-4x^2=0$

are

$x=\pm \frac{1}{2}$

Because we are looking for the section above the x-axis and the parabola “cries”, we get that the solution of the inequality is

$-\frac{1}{2}<x<\frac{1}{2}$

We intersect both results, meaning

$-1\geq x \text{  or  } x\geq 1$

and

$-\frac{1}{2}<x<\frac{1}{2}$

The intersection gives the empty group (there is no point sustaining the two inequalities).

Therefore, the final answer is that there is no real solution.

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