What is the domain of f(x)=sqrt(12x^2-7x+1) ?

The domain of f(x)=sqrt(12x^2-7x+1) is x = 1/3

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domain of f(x)=sqrt(12x^2-7x+1)

domain

f (x) = 12 x^{2} - 7 x + 1

x \leq \frac{1}{4} or x \geq \frac{1}{3}

Interval Notation

(- \infty, \frac{1}{4}] \cup [\frac{1}{3}, \infty)

Solution steps

Find non-negative values for radicals:

x \leq \frac{1}{4} or x \geq \frac{1}{3}

The function domain

x \leq \frac{1}{4} or x \geq \frac{1}{3}

f (x) = 2 x + 1, x > 3

u (x) = - 8 x + 16

f (x) = - \frac{1}{2 ( x + 1 )} + 3

f (x) = e^{a r c t a n (x)}

y, y \leq - x

What is the domain of f(x)=sqrt(12x^2-7x+1) ?
The domain of f(x)=sqrt(12x^2-7x+1) is x<= 1/4 \lor x>= 1/3