Question

a) Find the X- coordinates of all points -2π≤X≤2π, where the line y = x + b is tangent to the graph of f(x) = x + bsinx. b) Are the points of tangency described in part(a) relative maximum points of f ? Why? c) For all values ofstudent submitted b≻0 show that all inflection points of the graph of f lie on the line y = x.

Answer 1

a) The X-coordinates of the points of tangency are given by -b tan(π/4 - b/2) for -π/2 < btan(π/4 - b/2) < π/2.

b) These points are not relative maximums.

c) The inflection points are given by -π/2 + 2nπ ± arctan(-1/b) for n = 0, where b > 0. These points lie on the line y = x.

Step-by-step explanation:

In part (a), we're finding where the line y = x + b is tangent to the graph of f(x) = x + bsinx. To do this, we find where their slopes are equal. The slope of the line y = x + b is just 1, while the slope of f(x) is (1 + bsinx) - bcosx. Setting these slopes equal and solving for x gives us our desired X-coordinates. Note that this equation only has real solutions if certain conditions are met, as explained in part (b).

In part (b), we're checking whether these points are relative maximums or not. To do this, we use the first derivative test, which involves checking if the first derivative changes sign at a given point. If it does, then that point may be a relative maximum or minimum. In this case, our first derivative is (1 + bsinx) - bcosx, which changes sign at x = -b tan(π/4 + b/2). However, since our second derivative is always negative for all values of x and b, we know that these points cannot be relative maxima or minima. Instead, they're just points where our function changes direction from concave down to concave up or vice versa.

In part (c), we're finding where our function has inflection points. To do this, we need to find where its second derivative changes sign. This happens when sinx = -1/(b+1), which simplifies to x = -π/2 + 2nπ ± arctan(-1/(b+1)) for all values of n and any value of b > 0. Note that since our function only involves sine and cosine functions with no higher order terms involving these functions, it cannot have any inflection points other than those found here. These inflection points lie on the line y = x because they satisfy both f'(x) = 0 and f''(x) = 0 simultaneously at these values of x.

Answer 2

a) x = -π/2, π/2, 3π/2, -3π/2 (within the given range -2π ≤ x ≤ 2π)

b) No, the points of tangency are not relative maximum points.

a) Finding the X-coordinates of tangency points:

1. Set the slopes equal: The slope of the line y = x + b is 1. The slope of the graph of f(x) = x + bsinx is f'(x) = 1 + bcosx. For tangency, these slopes must be equal:

1 + bcosx = 1

2. Solve for x:

bcosx = 0

x = -π/2, π/2, 3π/2, -3π/2 (within the given range -2π ≤ x ≤ 2π)

b) Analyzing relative maximum points:

No, the points of tangency are not relative maximum points.

Explanation: The second derivative of f(x) is f''(x) = -bsinx.

At the points of tangency (x = -π/2, π/2, 3π/2, -3π/2), f''(x) = 0, indicating possible inflection points, not maxima.

c) Showing inflection points lie on y = x:

1. Set f''(x) = 0: -bsinx = 0

2. Solutions: x = nπ, where n is an integer (including the points of tangency).

3. Plug into f(x): f(nπ) = nπ + bsin(nπ) = nπ (since sin(nπ) = 0 for all integers n).

4. Conclusion: This shows that for any point of inflection (x, f(x)), we have f(x) = x, indicating that all inflection points lie on the line y = x.

Additional note: While I cannot incorporate images directly, I encourage you to visualize the graph of f(x) and the tangent lines to reinforce these concepts.