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

Question

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).

Answer 1

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?

Answer 2

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.