what is the solution of this inequality?: x+3/x^2-5x+4>_0?
To find the solution for the inequality \( \frac{{x+3}}{{x^2-5x+4}} \geq 0 \), we need to determine the values of \( x \) that satisfy the inequality.
First, let's find the critical points of the inequality. The critical points occur where the numerator or denominator is equal to zero. In this case, the denominator \( x^2-5x+4 \) can be factored as \( (x-1)(x-4) \). So, the critical points are \( x = 1 \) and \( x = 4 \).
Next, we create a number line with the critical points and test the intervals between them. We choose values from each interval and plug them into the inequality to determine if they satisfy the inequality.
For \( x < 1 \), we can choose a value like \( x = 0 \) and substitute it into the inequality:
\( \frac{{0+3}}{{0^2-5(0)+4}} = \frac{3}{4} > 0 \).
For \( 1 < x < 4 \), we can choose a value like \( x = 2 \) and substitute it into the inequality:
\( \frac{{2+3}}{{2^2-5(2)+4}} = \frac{5}{-2} < 0 \).
For \( x > 4 \), we can choose a value like \( x = 5 \) and substitute it into the inequality:
\( \frac{{5+3}}{{5^2-5(5)+4}} = \frac{8}{4} > 0 \).
Based on these tests, we can see that the inequality is satisfied for the intervals \( x < 1 \) and \( x > 4 \). Therefore, the solution to the inequality \( \frac{{x+3}}{{x^2-5x+4}} \geq 0 \) is \( x \in (-\infty, 1] \cup (4, \infty) \).