What is the domain of f(x)=sqrt(x^2-5x+1) ?

The domain of f(x)=sqrt(x^2-5x+1) is x = (sqrt(21)+5)/2

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domain of f(x)=sqrt(x^2-5x+1)

domain

f (x) = x^{2} - 5 x + 1

x \leq \frac{- 21 + 5}{2} or x \geq \frac{21 + 5}{2}

Interval Notation

(- \infty, \frac{- 21 + 5}{2}] \cup [\frac{21 + 5}{2}, \infty)

Solution steps

Find non-negative values for radicals:

x \leq \frac{- 21 + 5}{2} or x \geq \frac{21 + 5}{2}

The function domain

x \leq \frac{- 21 + 5}{2} or x \geq \frac{21 + 5}{2}

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

f (x) = (x + 1)^{\frac{1}{x}}

6 x - 12

y = 5^{x}

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

What is the domain of f(x)=sqrt(x^2-5x+1) ?
The domain of f(x)=sqrt(x^2-5x+1) is x<= (-sqrt(21)+5)/2 \lor x>= (sqrt(21)+5)/2