Suppose g(x) = 6x arctan (ln x/(x^3+2))

Plot y = g(x) in the square window defined by 0 ≤ x ≤ 2 and −1 ≤ y ≤ 1.
Also plot the secant lines connecting (1, g(1)) and (1 + h, g(1 + h)) for h = .5 and h = .25 in the same window.
Give a table of values of the slope of the secant lines connecting (1, g(1)) and (1 + 10^-j), g(1+10^-j)
when j is a positive integer ranging from 1 to 5. Using this information, guess what the derivative of g(x) at x = 1 would be, and guess the equation of the line tangent to
y = 6x arctan (ln x/(x^3+2))
�at (1, 0).

here's a table of values for

x,g(x), and secant slope:

1.10000000 0.18879484 1.88794837
1.01000000 0.01989861 1.98986135
1.00100000 0.00199900 1.99899859
1.00010000 0.00019999 1.99989999
1.00001000 0.00002000 1.99999000

I guess it's clear, eh?

visit http://rechneronline.de/function-graphs/

and enter

6x* arctan (log( x)/(x^3+2))
1.692*(x-1)
1.352 *(x-1)

for the three graph formulas, and set the x bounds for 0 to 2, and the y bounds for -1 to 1

plot the graphs, which are for g(x) and the two secants requested.

To plot the function and secant lines, you can follow these steps:

1. Determine the values of y = g(x) for the given range of x (0 ≤ x ≤ 2).
- Substitute each x-value into the function g(x) = 6x arctan (ln x/(x^3+2)) to calculate y.

2. Plot the graph of y = g(x) in the square window defined by 0 ≤ x ≤ 2 and −1 ≤ y ≤ 1.
- Use a graphing tool or software (e.g., Desmos, MATLAB, Excel) to create the plot.

3. Calculate the coordinates of the secant lines connecting (1, g(1)) and (1 + h, g(1 + h)) for h = 0.5 and h = 0.25.
- Substitute x = 1 and x = 1 + h into the function g(x) to find the y-coordinates for each secant line.

4. Plot the secant lines on the same graph as the function.
- Connect the points (1, g(1)) and (1 + h, g(1 + h)) with straight lines for each h value.

To determine the slope of the secant lines connecting (1, g(1)) and (1 + 10^(-j), g(1 + 10^(-j))), follow these steps:

1. Calculate g(1 + 10^(-j)) for j = 1 to 5.
- Substitute (1 + 10^(-j)) into the function g(x) = 6x arctan (ln x/(x^3+2)) to obtain the y-value.

2. Calculate the slope of each secant line using the formula: slope = (g(1 + 10^(-j)) - g(1))/(10^(-j)).
- Subtract g(1) from g(1 + 10^(-j)) and divide the result by 10^(-j).

3. Create a table with the values of j, g(1 + 10^(-j)), and the corresponding slope.

Using the information gathered from the table, you can make the following guesses:

1. The derivative of g(x) at x = 1 is the limit of the slopes as j approaches infinity.
- As j increases, the slope of the secant lines tends to a specific value, which represents the derivative at x = 1.

2. The equation of the tangent line at (1, 0) can be determined using the point-slope form.
- The slope of the tangent line is the derivative at x = 1, and the point (1, 0) lies on the line.

By following these steps and analyzing the data, you can approximate the derivative of g(x) at x = 1 and construct the equation of the tangent line at (1, 0).