The function g is defined for x>0 with g(1)=2, g'(x)=sin(x+1/x), and g"(x)=(1-1/x^2)cos(x+1/x).

A. Find all values of x in the interval 0.12<=x<=1 at which the graph of g has a horizontal tangent line.

B. On what subintervals of (0.12,1), if any, is the graph of g concave down? Justify you answer.

C. Write an equation for the line tangent to the graph of g at x=0.3.

D. Does the line tangent to the graph of g at x=0.3 lie above or below the graph of g for 0.3<x<1? Why?

I have no clue how to solve this. Can you please help me step by step. Thanks

http://dnichols30582.edublogs.org/files/2011/04/2010B2-2ibo364.pdf

Sure! Let's go through each part step by step.

A. To find the values of x in the interval [0.12, 1] where the graph of g has a horizontal tangent line, we need to find where the derivative of g, g'(x), equals zero. So we start by setting g'(x) = 0.

We are given g'(x) = sin(x + 1/x), so we have sin(x + 1/x) = 0.

To solve this, we need to find all values of x in the given interval that satisfy this equation.

To do this, we can use the fact that sin(x) = 0 for x = kπ, where k is an integer.
In our case, the equation becomes x + 1/x = kπ.

Now we solve for x by getting a common denominator:
x^2 + 1 = kπx

Rearranging the equation, we get:
x^2 - kπx + 1 = 0

This is a quadratic equation in terms of x. We can solve it using the quadratic formula:
x = (-(-kπ) ± √((-kπ)^2 - 4(1)(1)))/(2(1))

Simplifying further, we have:
x = (kπ ± √(k^2π^2 - 4))/2

Now we can substitute values of k from -2 to 2 and solve for x.

For each value of k, check if the resulting x falls within the interval [0.12, 1]. If it does, those values of x are where the graph of g has a horizontal tangent line in the given interval.

B. To determine where the graph of g is concave down, we need to examine the second derivative, g''(x).

Given g''(x) = (1 - 1/x^2)cos(x + 1/x), we can determine the intervals where g''(x) is negative, as in those intervals, the graph of g is concave down.

To find these intervals, set g''(x) less than zero:
(1 - 1/x^2)cos(x + 1/x) < 0

Now, we need to solve this inequality to find the values of x for which g''(x) < 0. The solution will give us the subintervals where g is concave down in the interval (0.12, 1).

C. To write an equation for the line tangent to the graph of g at x = 0.3, we need to find the value of g(0.3) and g'(0.3).

Given that g(1) = 2, we need to find the value of g(0.3) by integrating g'(x) from 1 to 0.3:
g(0.3) = g(1) + ∫[0.3, 1] g'(x) dx

To find g'(0.3), plug the value of x = 0.3 into g'(x):
g'(0.3) = sin(0.3 + 1/0.3)

Now, we have the value of g(0.3) and g'(0.3), we can use the point-slope form of a line to write the equation for the line tangent to the graph of g at x = 0.3.

D. To determine whether the tangent line at x = 0.3 lies above or below the graph of g for 0.3 < x < 1, we compare the values of the function g(x) at x = 0.3 and x = 1 to g'(0.3).

If g'(0.3) > g(x) for 0.3 < x < 1, then the tangent line lies above the graph. If g'(0.3) < g(x) for 0.3 < x < 1, then the tangent line lies below the graph.

By comparing the values, you can determine the relative position of the tangent line and the graph of g in this interval.