What is the domain of f(x)=sqrt(x^3-6x^2+8x) ?

The domain of f(x)=sqrt(x^3-6x^2+8x) is 0 = 4

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domain of f(x)=sqrt(x^3-6x^2+8x)

domain

f (x) = x^{3} - 6 x^{2} + 8 x

0 \leq x \leq 2 or x \geq 4

Interval Notation

[0, 2] \cup [4, \infty)

Solution steps

Find non-negative values for radicals:

0 \leq x \leq 2 or x \geq 4

The function domain

0 \leq x \leq 2 or x \geq 4

f (x) = x^{3} - 12 x + 1

\frac{4 x + 9}{3 x - 2}

x + 2 y = 16

x^{2} - 2 x - 35

f (x) = \frac{9}{x} + x + 1

What is the domain of f(x)=sqrt(x^3-6x^2+8x) ?
The domain of f(x)=sqrt(x^3-6x^2+8x) is 0<= x<= 2\lor x>= 4