What is the domain of f(x)=sqrt(2x^2-11x+12+1) ?

The domain of f(x)=sqrt(2x^2-11x+12+1) is x = (sqrt(17)+11)/4

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domain of f(x)=sqrt(2x^2-11x+12+1)

domain

f (x) = 2 x^{2} - 11 x + 12 + 1

x \leq \frac{- 17 + 11}{4} or x \geq \frac{17 + 11}{4}

Interval Notation

(- \infty, \frac{- 17 + 11}{4}] \cup [\frac{17 + 11}{4}, \infty)

Solution steps

Find non-negative values for radicals:

x \leq \frac{- 17 + 11}{4} or x \geq \frac{17 + 11}{4}

The function domain

x \leq \frac{- 17 + 11}{4} or x \geq \frac{17 + 11}{4}

\frac{7 x}{2 x + 5}

x + y = 1

\frac{x + 1}{x + 1}

2^{x} + 1

f (x) = ln (- x^{2} + 3 x - 2)

What is the domain of f(x)=sqrt(2x^2-11x+12+1) ?
The domain of f(x)=sqrt(2x^2-11x+12+1) is x<= (-sqrt(17)+11)/4 \lor x>= (sqrt(17)+11)/4