derivatives
Calculus Techniques and solving limits


Continuity and Discontinuity
Infinite Limits and Asymptotes
IROC AND AROC
100

x^2

2x

100

What are the 5 ways to evaluate limit? 

1) direct substitution

2) Rationalizing 

3) factoring 

4) Using trig 

5) value of table 

6) L'Hôpital's Rule

100

If a function f(x) is continuous at the point x = 3 and f(3) = 7, what is the value of the limit lim(x -> 3) f(x)?

also 7

100

What is the vertical asymptote of the function f(x) = 1/(x - 2)?

 vertical asymptote of the function is x = 2.

100

s(t)=3t^3−4t^2+2t−1 meters at time t in seconds. Find the instantaneous rate of change of the position with respect to time at t=2.

s′(t)=9t2−8t+2

s′(2):22 m/s

200

2x^5+5x^3

10x^4+15x^2

200

What are some common types of discontinuities in functions?

discontinuities include removable, jump, and infinite discontinuities.

200

Consider a function g(x) defined over the interval [1, 5]. If g(1) = 2 and g(5) = 8, and g(x) is continuous over the entire interval, what is the average rate of change of the function over this interval?

 (g(5) - g(1))/(5 - 1) = (8 - 2)/4 = 6/4 = 3/2 or 1.5.

200

Determine the limit as x approaches 3 for the function g(x) = (x^2 - 9)/(x - 3).

The limit as x approaches 3 for the function g(x) = (x^2 - 9)/(x - 3) is 6.

200

v(t)=5t.  Find the instantaneous rate of change of velocity at t=4 seconds.

take the derivative so 5

300

(x^2+3x)/(x-1)

(x^2-x-3)/(x-1)^2

300

What does the Squeeze Theorem state, and how can it be used to determine limits? (traditional way)

The Squeeze Theorem states that if two functions are on top of each other, and the two bounding functions have the same limit as the target function, then the target function also has that limit.

300
  • Given a function g(x) = (x^2 - 4)/(x - 2), what value should be assigned to g(2) to remove the discontinuity at x = 2? 

To remove the discontinuity at x = 2 for the function g(x) = (x^2 - 4)/(x - 2), we should assign the value g(2) = 4.

300

For the function h(x) = (3x^2 + 2x - 1)/(2x^2 + x - 5), find the horizontal asymptote as x approaches infinity.

The horizontal asymptote of the function h(x) = (3x^2 + 2x - 1)/(2x^2 + x - 5) as x approaches infinity is y = 3/2.

300

A bicycle travels 15 miles in 1.5 hours. Find the average speed of the bicycle.

total distance / total time = 10 

400

 sin(2x^3+1)

6x^2cos(2x^3+1)

400

Solve Limit as x approaches 0 of (3x)/(x)

3

400

What are the three ways checking for continuity?

1) f(c) exists

2) limit as x approaches c of f(x) exists 

3) if number 1 equal to number 2 

400

Calculate the average rate of change of the function h(t) = 2t^2 - 3t + 1 over the interval [1, 2].

The average rate of change of h(t) = 2t^2 - 3t + 1 over the interval [1, 2] is 3.


400

bicycle A travels 15 miles in 1.5 hours. Bike B travels 20 miles in 3 hours. Find the average speed of the bicycle.

bike a = 10 

Bike b= 6.67

500

(x^2+1)(3x-2)

9x^2-4x+3
500

Calculate the limit as x approaches 2 of the function of (x^2-4)/(x-2)

4

500

To confirm continuity over an interval, what conditions must a function satisfy, and how can we check for continuity on a closed interval?

  1. The function is defined on [a, b].
  2. The function is continuous at every point within the interval (a, b).
  3. The limits of the function as 'x' approaches 'a' and 'b' exist.
500

Calculate the limit as x approaches negative infinity for the function k(x) = (4x^3 + 2x^2 - 3)/(3x^3 - 5x + 1).

4/3.

500

A train travels a distance of 300 miles. It takes 4.5 hours to complete the journey. However, during the first 2 hours, it travels at a constant speed of 60 miles per hour, and during the remaining time, it travels at a constant speed of 80 miles per hour. Find the average speed of the train for the entire journey.

1) convert units 

For the first 2 hours at 60 mph, the distance covered is 60 mph×2 hours=120 miles60mph×2hours=120miles.

For the remaining 2.5 hours at 80 mph, the distance covered is 80 mph×2.5 hours=200 miles80mph×2.5hours=200miles.

and then calculate time and distance 

320 miles / 4.5 hours = 71.11 miles per hour