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.

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

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

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

22. Limits & Continuity

Continuity

Precalculus

22. Limits & Continuity

Continuity

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Multiple Choice

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

$x=-4,x=3$

$x=4,x=-3$

$x=4$

Function is continuous everywhere.

Verified step by step guidance

Step 1: Identify the type of function given. The function f(x) = (x - 4) / (x^2 - x - 12) is a rational function, which is a ratio of two polynomials.

Step 2: Determine where the function might be discontinuous. Rational functions are discontinuous where the denominator is equal to zero, as division by zero is undefined.

Step 3: Set the denominator equal to zero and solve for x. The denominator is x^2 - x - 12. Set this equal to zero: x^2 - x - 12 = 0.

Step 4: Factor the quadratic equation x^2 - x - 12 = 0. Look for two numbers that multiply to -12 and add to -1. The factors are (x - 4)(x + 3) = 0.

Step 5: Solve the factored equation for x. Set each factor equal to zero: x - 4 = 0 and x + 3 = 0. Solving these gives x = 4 and x = -3. These are the values where the function is discontinuous.

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Struggling with Precalculus?

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​

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