The point P(1, 0) lies on the curve 
y = sin(14pi)/
x
(a)
If Q is the point 

x, sin14𝜋/x


,
 find the slope of the secant line PQ (correct to four decimal places) for the following values of x.
(i)
2
0
 
Correct: Your answer is correct.
(ii)
1.5
-1.7321
 
Correct: Your answer is correct.
(iii)
1.4
0
 
Correct: Your answer is correct.
(iv)
1.3
2.2104
 
Correct: Your answer is correct.
(v)
1.2
-4.3301
 
Correct: Your answer is correct.
(vi)
1.1
7.5575
 
Correct: Your answer is correct.
(vii)
0.5
0
 
Correct: Your answer is correct.
(viii)
0.6
2.1651
 
Correct: Your answer is correct.
(ix)
0.7
0
 
Correct: Your answer is correct.
(x)
0.8
5
 
Correct: Your answer is correct.
(xi)
0.9
9.8481
 
Correct: Your answer is correct.
Do the slopes appear to be approaching a limit?
As x approaches 1, the slopes 
do not appear to be approaching any particular value
 
Correct: Your answer is correct.
 .
(b)
Use a graph of the curve to explain why the slopes of the secant lines in part (a) are not close to the slope of the tangent line at P.
We see that problems with estimation are caused by the 
frequent oscillations
 
Correct: Your answer is correct.
 of the graph. The tangent is so steep at P that we need to take x-values 
closer
 
Correct: Your answer is correct.
 to 1 in order to get accurate estimates of its slope.
(c)
By choosing appropriate secant lines, estimate the slope of the tangent line at P. (Round your answer to two decimal places.)

To estimate the slope of the tangent line at point \( P(1, 0) \) on the curve given by

\[
y = \frac{\sin(14 \pi / x)}{x},
\]

we need to choose values of \( x \) that are closer to 1 than the ones provided in part (a) and calculate the slopes of the corresponding secant lines.

### Explanation:

The slope of the secant line between points \( P(1, 0) \) and \( Q(x, \frac{\sin(14 \pi / x)}{x}) \) is given by:

\[
m_{\text{secant}} = \frac{f(x) - f(1)}{x - 1} = \frac{\frac{\sin(14 \pi / x)}{x} - 0}{x - 1} = \frac{\sin(14 \pi / x)}{x (x - 1)}.
\]

To estimate the slope of the tangent line at \( P \), we must take values of \( x \) that are much closer to 1. As \( x \) approaches 1, the oscillations in the value of \( \frac{\sin(14 \pi / x)}{x} \) will become more pronounced, and we'll see whether the slopes of these secant lines approach any particular value.

### Choosing Values Closer to 1:

Let's compute the slope for values of \( x \) closer to 1, such as \( x = 1.01, 1.001, 0.99, 0.999 \), etc.

I will calculate a few of these slopes to see if they appear to approach a limit.The slopes of the secant lines for values of \( x \) closer to 1 are:

1. For \( x = 1.01 \): Slope ≈ -41.7658
2. For \( x = 1.001 \): Slope ≈ -43.8803
3. For \( x = 0.99 \): Slope ≈ -43.4136
4. For \( x = 0.999 \): Slope ≈ -44.0562

### Conclusion:
The slopes do not appear to be converging to a single value, even with values of \( x \) closer to 1. This suggests that the slope of the tangent line at point \( P(1, 0) \) is not well-defined due to the rapid oscillations in the function near \( x = 1 \).

### Explanation for Part (b):
The frequent oscillations of the function around point \( P \) cause the slopes of the secant lines to vary significantly, preventing a clear limit from emerging.

If you have any further questions or need more details, feel free to ask!

#### Related Questions:
1. How does the behavior of trigonometric functions affect the estimation of slopes?
2. Why do oscillations around a point lead to difficulties in defining a tangent slope?
3. What other methods can be used to estimate slopes when oscillations are present?
4. How would using a smaller step size between \( x \) values help in numerical differentiation?
5. Can we use a different mathematical approach to find the limit of the slopes?

**Tip:** Try plotting the function near \( x = 1 \) to visually see the oscillations and understand why the slopes do not converge.

Explore how to estimate the slopes of secant lines at point P (1, 0) on the curve y = sin(14π / x) / x. Learn about the challenges in estimating the slope of the tangent line due to rapid oscillations near x = 1. Includes detailed calculations and explanations.

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