fbpx
calculus online - Free exercises and solutions to help you succeed!

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

There is no real solution

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.

Have a question? Found a mistake? – Write a comment below!
Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions! 

Share with Friends

Leave a Reply