What is the domain of f(x)=(10x^2-9)^{1/4}-x ?

The domain of f(x)=(10x^2-9)^{1/4}-x is x = 3/(10^{1/2)}

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domain of f(x)=(10x^2-9)^{1/4}-x

domain

f (x) = (10 x^{2} - 9)^{\frac{1}{4}} - x

x \leq - \frac{3}{1 0 ^{\frac{1}{2}}} or x \geq \frac{3}{1 0 ^{\frac{1}{2}}}

Interval Notation

(- \infty, - \frac{3}{1 0 ^{\frac{1}{2}}}] \cup [\frac{3}{1 0 ^{\frac{1}{2}}}, \infty)

Solution steps

Find non-negative values for radicals:

x \leq - \frac{3}{1 0 ^{\frac{1}{2}}} or x \geq \frac{3}{1 0 ^{\frac{1}{2}}}

The function domain

x \leq - \frac{3}{1 0 ^{\frac{1}{2}}} or x \geq \frac{3}{1 0 ^{\frac{1}{2}}}

h (x) = \frac{2 x + 6}{5}

f (x) = \frac{x ^{2} + 8 x - 9}{x ^{2} + 3 x - 4}

\frac{x - 1}{x ^{2}}

3 - 4 sin (\frac{2}{3} (x - 1))

f (x) = \frac{5}{x}

What is the domain of f(x)=(10x^2-9)^{1/4}-x ?
The domain of f(x)=(10x^2-9)^{1/4}-x is x<=-3/(10^{1/2)}\lor x>= 3/(10^{1/2)}