 |
 |
 |
 |
|
|
|
|
2.9, #2, Given Problem.
 |
2.9, #2, Given Problem. |
 |
Part (a).
Plot the function. If you were plotting the curve in Maple, >plot(ln(x),x=0..5,y=-2..2); Along with this plot there is a sketch of a line tangent to the curve at (1,0). It appears that the slope of the tangent line is approximately 1. |
 |
Part (b).
Using the slope-intercept form of a line and our
approximate slope (m)
of 1, we can get the linear equation for the tangent line at (1,0).
|
 |
Part (c).
Here we use our linear approximation to find approximations for f (0.9) and f ( 1.3). The estimates must be slightly high, because our tangent line is above the curve of the function. |
2.9, #8, Given Problem.
 |
2.9, #8, Given Problem. |
 |
Here we have a summary of the given information. We are to find an estimate for P when h = 3. |
 |
We calculate the average rate of change (slope) between the h = 1 and h = 2 data points. |
 |
Now using our approximate slope, we substitute in the latest data point (2,74.9) and find the y-intercept (b). |
 |
Now we have our linear approximation equation. |
 |
Using this equation, we can find out an approximation for when h = 3. |
 |
2.9, #14, Given Problem. |
 |
Part (a).
This table summarizes the given information, and shows the two values we want to find. I found the slope of 3 by using the g '(x) formula which was given. |
 |
We use our slope and substitute in our known data point of (2,-4) to find b. |
 |
Now we have our linear approximation model. |
 |
Using our model, we find approximations for g(1.95) and g(2.05). |
|
|
We know this because from our g'(x) formula we can see that as x increases, g'(x) increases. This means the slope of our missing function is increasing (or concave up). Therefore, our linear estimate must be below the graph. |
|
|
|
|
|
|
|
|
|
