What is the domain of f(x)=(sqrt(x))/(8x^2+7x-1) ?

The domain of f(x)=(sqrt(x))/(8x^2+7x-1) is 0 1/8

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domain of f(x)=(sqrt(x))/(8x^2+7x-1)

domain

f (x) = \frac{x}{8 x ^{2} + 7 x - 1}

0 \leq x < \frac{1}{8} or x > \frac{1}{8}

Interval Notation

[0, \frac{1}{8}) \cup (\frac{1}{8}, \infty)

Solution steps

Find non-negative values for radicals:

x \geq 0

Find undefined (singularity) points:

x = \frac{1}{8}, x = - 1

Combine real regions and undefined points for final function domain

0 \leq x < \frac{1}{8} or x > \frac{1}{8}

y = x + 2

f (x) = 2 x e^{- x^{2}}

f (x) = 1 0^{1 - x^{3}}

f (x) = \frac{( x - 6 )}{8}

f (x) = e^{- 0.5 x^{2}}

What is the domain of f(x)=(sqrt(x))/(8x^2+7x-1) ?
The domain of f(x)=(sqrt(x))/(8x^2+7x-1) is 0<= x< 1/8 \lor x> 1/8