Use graph of to determine if the function is continuous or discontinuous at
Table of contents
- 0. Functions7h 55m
- Introduction to Functions18m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms36m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals4h 44m
- 9. Graphical Applications of Integrals2h 27m
- 10. Physics Applications of Integrals 3h 16m
- 11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 31m
- 12. Techniques of Integration7h 41m
- 13. Intro to Differential Equations2h 55m
- 14. Sequences & Series5h 36m
- 15. Power Series2h 19m
- 16. Parametric Equations & Polar Coordinates7h 58m
1. Limits and Continuity
Continuity
Multiple Choice
Determine the value(s) of x (if any) for which the function is discontinuous.
f(x)=x2−x−12x−4
A
x=−4,x=3
B
x=4,x=−3
C
D
Function is continuous everywhere.
3 Comments
Verified step by step guidance1
Identify the type of function: The given function is a rational function, which is a fraction where both the numerator and the denominator are polynomials.
Determine where the function is undefined: A rational function is undefined where its denominator is zero. Set the denominator equal to zero and solve for x: \(x^2 - x - 12 = 0\).
Factor the quadratic equation: The equation \(x^2 - x - 12 = 0\) can be factored into \((x - 4)(x + 3) = 0\).
Find the roots of the factored equation: Set each factor equal to zero and solve for x. This gives \(x - 4 = 0\) which results in \(x = 4\), and \(x + 3 = 0\) which results in \(x = -3\).
Conclude the points of discontinuity: The function is discontinuous at the values of x where the denominator is zero, which are \(x = 4\) and \(x = -3\).
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