Question

Find the domain of the real valued function: f(x) = $\sqrt{ log_{0.3} (x - x^{2})}$

Answer 1

Answer:

Thus the domain of the function is 0 < x < 1.

Step-by-step explanation:

To find the domain of the real valued function:

$f(x)=\sqrt{ log_{0.3} (x - x^{2})}$

as we know that square root does not defined for negative values

so

$log_{a}(x^{b}) = b.log_{a}x \\\\\\log_{\frac{3}{10} }(\sqrt{x-x^{2} } ) = \frac{1}{2} .log_{\frac{3}{10} }}(x-x^{2})\\\\ \frac{1}{2} .log_{\frac{3}{10} }}(x-x^{2})\geq 0 \\\\ or\\\\x(1-x) \geq 0\\\\\\0 < x < 1$

Thus the domain of the function is 0 < x < 1.

Answer 2

Thus the domain of the function is 0 < x < 1.

Step-by-step explanation:

To find the domain of the real valued function:

\begin{lgathered}f(x)=\sqrt{ log_{0.3} (x - x^{2})}\\\\\end{lgathered}f(x)=log0.3​(x−x2)​​

as we know that square root does not defined for negative values

so



Thus the domain of the function is 0 < x < 1.