# How do you find the domain for f(x) = sqrt(4 - x^5)?

Apr 9, 2015

Assuming we are restricted to Real numbers (i.e. $f \left(x\right) \epsilon \mathbb{R}$)
Then the domain of $f \left(x\right) = \sqrt{4 - {x}^{5}}$
is all values of $x$ for which the argument of the square root is non-negative.
That is
${x}^{5} \le 4$
or
$x \le \sqrt[5]{4}$
So the domain of $f \left(x\right)$ is $\left(- \infty , \sqrt[5]{4}\right]$

Apr 9, 2015

Domain: All real numbers $\le$ (4)$^ \frac{1}{5}$

The domain of f(x) would be all real values of x for which 4-${x}^{5}$ is greater or equal to 0. The values of x which make 4-${x}^{5}$ negative cannot be accepted because in that case $\sqrt{4 - {x}^{5}}$ would become imaginary.
Hence solving the inequality 4-${x}^{5}$ $\ge$ 0, it would be 4$\ge$ ${x}^{5}$. That is 4^$\frac{1}{5}$ $\ge$ x

In the interval notation it would be ( -inf, 4^$\frac{1}{5}$]