Foundations of Calculus Practice Test

Review the following concepts if needed:

• Limits Help
• Properties of Limits Help
• Continuity Help
• The Derivative Help
• Rules for Calculating Derivatives Help
• The Derivative as a Rate of Change Help

Foundations of Calculus Practice Test

1. Calculate, if possible, each of these limits. Give reasons for each step of your solution.

   

2. Determine whether the given function f is continuous at the given point c . Give careful justifications for your answers.

   

3. Use the definition of derivative to calculate each of these derivatives.

   (a) f′ (2) when f(x) = x ² + 4 x

   (b) f′ (1) when f(x) = −1/ x ²

4. Calculate each of these derivatives. Justify each step of your calculation.

   

   

5. Imitate the example in the text to do each of these falling body problems.

   (a) A ball is dropped from a height of 100 feet. How long will it take that ball to hit the ground?

   (b) Suppose that the ball from part (a) is thrown straight down with an initial velocity of 10 feet per second. Then how long will it take the ball to hit the ground?

   (c) Suppose that the ball from part (a) is thrown straight up with an initial velocity of 10 feet per second. Then how long will it take the ball to hit the ground?

6. Use the Chain Rule to perform each of these differentiations:

   

7. If a car has position p(t) = 6 t ² − 5 t + 20 feet, where t is measured in seconds, then what is the velocity of that car at time t = 4? What is the average velocity of that car from t = 2 to t = 8? What is the greatest velocity over the time interval [5, 10]?

8. In each of these problems, use the formula for the derivative of an inverse function to find [ f ⁻¹ ]′(1).

   (a) f (0) = 1, f′ (0) = 3

   (b) f (3) = 1, f′ (3) = 8

   (c) f (2) = 1, f′ (2) = ²

   (d) f (1) = 1, f′ (1) = 40

Answers

1.  

   

   

2.  

   

3. (a) We calculate

   

   The derivative is therefore equal to 8.

   (b) We calculate

   

   The derivative is therefore equal to 2.

4.  

   

5. (a) Since the ball is dropped, v ₀ = 0. The initial height is h ₀ = 100. Therefore the position of the body at time t is given by

   p ( t ) = −16 t ² + 0 · t + 100.

   The body hits the ground when

   0 = p ( t ) = 16 t ² + 100

   or t = 2.5 seconds.

   (b) Since the ball has initial velocity 10 feet/second straight down, we know that v ₀ = −10. The initial height is h ₀ = 100. Therefore the position of the body at time t is given by

   p ( t ) = −16 t ² − 10 · t + 100.

   The body hits the ground when

   0 = p ( t ) = −16 t ² − 10 t + 100

   or t ≈ 2.207 seconds.

   (c) Since the ball has initial velocity 10 feet/second straight up, we know that v ₀ = 10. The initial height is h ₀ = 100. Therefore the position of the body at time t is given by

   p ( t ) = −16 t ² + 10 · t + 100.

   The body hits the ground when

   0 = p ( t ) = −16 t ² + 10 t + 100

   or t ≈ 2.832 seconds.

6.  

   

7. Of course v ( t ) = p ′( t ) = 12 t − 5 so v (4) = 43 feet/second. The average velocity from t = 2 to t = 8 is

   

   The derivative of the velocity function is ( v ′)′ ( t ) = 12. This derivative never vanishes, so the extrema of the velocity function on the interval [5, 10] occur at t = 5 and t = 10. Since v (5) = 55 and v (10) = 115, we see that the maximum velocity on this time interval is 115 feet per second at t = 10.

8. (a) We know that

   

   (b) We know that

   

   (c) We know that

   

   (d) We know that