Question

week 2 group project

the police department has a machine that makes a graph for each car that drives on a certain road, graphing the position of the car (in miles) as a function of time (in minutes).

1. suppose the six graphs from the week 1 group project are graphs that this machine recorded.

for cars (a), (b), and (c) from week 1, calculate the average velocity on the interval 2,3, i.e. between 2 and 3 minutes. then calculate the average velocity over 2,2.5. which calculation seems closer to the instantaneous velocity at 2 minutes?

1. one way of estimating the instantaneous velocity at 2 minutes is by calculating average velocities over 2,2 + h, and choosing smaller and smaller values for the interval length h. you started this process already in question #1. (h was 1 for the interval 2,3 and .5 for the interval 2,2.5.) continue to carry out this strategy for one more smaller interval on cars (a), (b), and (c). use the results from questions #1 and #2 to approximate what the instantaneous velocity at 2 minutes is for each car.

1. create a graph of a position function where the average velocity over 2,3 is a better estimate for the instantaneous velocity at 2 minutes than the average velocity over 2,2.5.

1. determine whether the following statements are true and give an explanation or counterexample.

a. the value of \\(\lim_{x \to 3} \frac{x^2 - 9}{x - 3}\\) does not exist.
b. the value of \\(\lim_{x \to a} f(x)\\) can always be found by computing \\(f(a)\\).
c. the value of \\(\lim_{x \to a} f(x)\\) does not exist if \\(f(a)\\) is undefined.

Explanation:

To solve part 4a, we analyze the limit:

Step 1: Simplify the function

We have the function \( \frac{x^2 - 9}{x - 3} \). Notice that \( x^2 - 9 \) is a difference of squares, so we can factor it as \( (x - 3)(x + 3) \). So the function becomes:
\[
\frac{(x - 3)(x + 3)}{x - 3}
\]
For \( x
eq 3 \), we can cancel out the \( x - 3 \) terms (since \( x = 3 \) makes the original function undefined, but we are looking at the limit as \( x \) approaches 3, not the value at \( x = 3 \)). After canceling, we get \( x + 3 \).

Step 2: Evaluate the limit

Now we find the limit as \( x \) approaches 3 of \( x + 3 \). Using the direct substitution property (since \( x + 3 \) is a polynomial, which is continuous everywhere), we substitute \( x = 3 \) into \( x + 3 \):
\[
\lim_{x \to 3} (x + 3) = 3 + 3 = 6
\]
So the limit exists and is equal to 6. Therefore, the statement "The value of \( \lim_{x \to 3} \frac{x^2 - 9}{x - 3} \) does not exist" is false.

This statement is false. The limit of a function as \( x \) approaches \( a \) depends on the behavior of the function near \( x = a \), not necessarily the value at \( x = a \). For example, consider the function \( f(x) = \frac{x^2 - 9}{x - 3} \) (from part 4a). At \( x = 3 \), \( f(3) \) is undefined (since we get \( \frac{0}{0} \)). However, as we saw in part 4a, \( \lim_{x \to 3} f(x) = 6 \). Another example: consider a piece - wise function \( f(x)=

$$\begin{cases}x + 1, & x eq2\\5, & x = 2\end{cases}$$

\). Then \( \lim_{x \to 2} f(x)=\lim_{x \to 2}(x + 1)=3\), but \( f(2) = 5 \). So we cannot always find the limit as \( x\to a \) by computing \( f(a) \).

This statement is false. Using the same example as in part 4a, \( f(x)=\frac{x^2 - 9}{x - 3} \) is undefined at \( x = 3 \) (since \( f(3)=\frac{3^2-9}{3 - 3}=\frac{0}{0} \), which is undefined). But we found that \( \lim_{x \to 3}f(x) = 6 \), so the limit can exist even if the function is undefined at the point \( x = a \).

Answer:

(for part 4a)
The statement is false. The limit \( \lim_{x \to 3} \frac{x^2 - 9}{x - 3} \) exists and is equal to 6 (we simplify \( \frac{x^2 - 9}{x - 3} \) to \( x + 3 \) for \( x
eq 3 \), then substitute \( x = 3 \) to get 6).

For part 4b, we analyze the statement "The value of \( \lim_{x \to a} f(x) \) can always be found by computing \( f(a) \)".