Practice Problems
Site: | Saylor Academy |
Course: | MA005: Calculus I |
Book: | Practice Problems |
Printed by: | Guest user |
Date: | Tuesday, October 22, 2024, 1:50 AM |
Description
Work through the odd-numbered problems 1-17. Once you have completed the problem set, check your answers.
1. Use the function in Fig. 12 to fill in the table and then graph .
= the estimated slope of the tangent line to at the point (x,y) | ||
---|---|---|
0 | ||
0.5 | ||
1.0 | ||
1.5 | ||
2.0 | ||
2.5 | ||
3.0 | ||
3.5 | ||
4.0 |
3.
(a) At what values of does the graph of f in Fig. 14 have a horizontal tangent line?
(b) At what value(s) of is the value of f the largest? smallest?
(c) Sketch the graph of = the slope of the line tangent to the graph of at the point .
5.
(b) Sketch the graph of = slope of the line tangent to the graph of at the point .
(c) Your graph in part (b) should look familiar. What function is it?
Problems 7 – 9 assume that a rocket is following the path , from left to right.
7. At what point should the engine be turned off in order to coast along the tangent line to a base at ?
9. At what point should the engine be turned off in order to coast along the tangent line to a base at ?
For each function in problems 11-15, perform steps :
In problem 17, use the result that if then .
17. At which point(s) does the line tangent to the graph at that point also go through the point ?
Source: Dale Hoffman, https://s3.amazonaws.com/saylordotorg-resources/wwwresources/site/wp-content/uploads/2012/12/MA005-3.1-Introduction-to-Derivatives.pdf
This work is licensed under a Creative Commons Attribution 3.0 License.
1.
|
= the estimated slope of the tangent line to at the point | |
---|---|---|
0 | 1 | 1 |
0.5 | 1.4 | 1/2 |
1.0 | 1.6 | 0 |
1.5 | 1.4 | –1/2 |
2.0 | 1 | –2 |
2.5 | 0 | –2 |
3.0 | –1 | –2 |
3.5 | –1.3 | 0 |
4.0 | –1 | 1 |
3.
7. The solution is similar to the method used in Example 4. Assume we turn off the engine at the point on the curve , and then find values of and so the tangent line to at the point goes through the given point is on so . The equation of the tangent line to at is so, substituting and , we have Solving we get or . The solution we want (moving left to right along the curve) is ( would be the solution if we were moving right to left).
9. Impossible. The point is "inside" the parabola.
11.
13.
15.
17. , so . The problem is to find for which
This reduces to so or and the required points are and .