Exercise
Determine the domain of the function:
y=\sqrt{x^2-1}+\ln\sqrt{1-4x^2}
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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