What is the domain of f(x)=sqrt(x^2-4x)+(x^2-3x+2) ?

The domain of f(x)=sqrt(x^2-4x)+(x^2-3x+2) is x = 4

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domain of f(x)=sqrt(x^2-4x)+(x^2-3x+2)

domain

f (x) = x^{2} - 4 x + (x^{2} - 3 x + 2)

x \leq 0 or x \geq 4

Interval Notation

(- \infty, 0] \cup [4, \infty)

Solution steps

Find non-negative values for radicals:

x \leq 0 or x \geq 4

The function domain

x \leq 0 or x \geq 4

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

\frac{ln ( x )}{ln ( 3 )}

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

g (x) = 2 x^{3} - 6 x + 3, - 2 \leq x \leq 2

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

What is the domain of f(x)=sqrt(x^2-4x)+(x^2-3x+2) ?
The domain of f(x)=sqrt(x^2-4x)+(x^2-3x+2) is x<= 0\lor x>= 4