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

The domain of f(x)=sqrt(-x^2+4x-3) is 1<= x<= 3

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

domain

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

1 \leq x \leq 3

Interval Notation

[1, 3]

Solution steps

Find non-negative values for radicals:

1 \leq x \leq 3

The function domain

1 \leq x \leq 3

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

3 x + 2

f (x) = - 2 - x

f (x) = ∣ x + 6 ∣

f (x) = \frac{5}{x + 5}

What is the domain of f(x)=sqrt(-x^2+4x-3) ?
The domain of f(x)=sqrt(-x^2+4x-3) is 1<= x<= 3