What is the domain of f(x)=(sqrt(2x-7))/(x^2-5x+4) ?

The domain of f(x)=(sqrt(2x-7))/(x^2-5x+4) is 7/2 4

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domain of f(x)=(sqrt(2x-7))/(x^2-5x+4)

domain

f (x) = \frac{2 x - 7}{x ^{2} - 5 x + 4}

\frac{7}{2} \leq x < 4 or x > 4

Interval Notation

[\frac{7}{2}, 4) \cup (4, \infty)

Solution steps

Find non-negative values for radicals:

x \geq \frac{7}{2}

Find undefined (singularity) points:

x = 4, x = 1

Combine real regions and undefined points for final function domain

\frac{7}{2} \leq x < 4 or x > 4

f (x) = x^{2} e^{x}

g (x) = x + 1

\frac{1}{( y ^{2} + 1 ) ^{\frac{1}{2}}} y

4 x + 7

f (x) = lo g_{3} (x + 3)

What is the domain of f(x)=(sqrt(2x-7))/(x^2-5x+4) ?
The domain of f(x)=(sqrt(2x-7))/(x^2-5x+4) is 7/2 <= x<4\lor x>4