Question

solve the inequality \\(\dfrac{-8}{(x + 3)(x - 11)} \geq 0\\). enter the solution in interval notation.

Explanation:

Step1: Find critical points

Set the denominator equal to zero: \((x + 3)(x - 11)=0\). So \(x=-3\) or \(x = 11\) are critical points. Also, note the numerator is \(-8\) (a constant).

Step2: Analyze sign of the function

The function is \(\frac{-8}{(x + 3)(x - 11)}\). The numerator \(-8\) is negative. We need the fraction to be non - negative (greater than or equal to zero). A negative divided by a number is non - negative when the denominator is negative (since negative/negative = positive).
So we solve \((x + 3)(x - 11)<0\) (because numerator is negative, for \(\frac{-8}{(x + 3)(x - 11)}\geq0\), denominator must be negative).
The quadratic function \(y=(x + 3)(x - 11)=x^{2}-8x - 33\) is a parabola opening upwards (since the coefficient of \(x^{2}\) is positive). The roots are \(x=-3\) and \(x = 11\). The function is negative between the roots. So \(-3We also need to check the points where the denominator is zero (\(x=-3\) and \(x = 11\)): at \(x=-3\) and \(x = 11\), the denominator is zero, so the function is undefined at these points, so they are not included in the solution.

Answer:

\((-3, 11)\)