Determine the value(s) of xxx (if any) for which the function is discontinuous.f(x)=x−4x2−x−12f\left(x\right)=\frac{x-4}{x^2-x-12}f(x)=x2−x−12x−4

x = 4 , x = − 3 x=4,x=-3 x = 4 , x = − 3

Determine the value(s) of x x x (if any) for which the function is discontinuous. f ( x ) = x − 4 x 2 − x − 12 f\left(x\right)=\frac{x-4}{x^2-x-12} f ( x ) = x 2 − x − 12 x − 4

In this problem, we're asked to determine the values of x, if any, for which this function is discontinuous. Now the function that we're given here is f of x x is equal to x minus 4 over x squared minus x minus 12. Now remember when working with a rational function, determining where this function is discontinuous means that I need to take that denominator and set it equal to 0. So setting my denominator equal to 0 here gives me a quadratic equation to solve. So the easiest thing to do here is going to be to go ahead and factor this. Now I wanna look for two numbers that multiply to negative 12 and add to negative one. So this factors to x minus 4 times x plus 3. Now since this is equal to 0, that means that I can go ahead and take each of these individual terms and set them equal to 0. So if I take x minus 4 equals 0 and x plus 3 equals 0 and solve each of these equations, I'll get my values of x for which this function is discontinuous. So if I add 4 on both sides for this equation here, I end up with x equals 4. Then over here, if I subtract 3 on both sides, I end up with x equals negative 3. So here I have my two values for which this function is discontinuous, x equals 4 and x equals negative 3. Now if you're curious what type of discontinuities we're dealing with here, we can look back to our original function, and I see that my numerator is x minus 4. Now since one of my factors in my denominator was also x minus 4, this represents a removable discontinuity or a hole at x equals 4. Then my or a hole at x equals 4. Then my term x or my solution x equals negative 3, where it is discontinuous. This represents an asymptote. Let me know if you have any questions here, and I'll see you in the next video.