Question

Determine on what intervals (if any) the graph of f(x) = 2 sin x - cos 2x is concave up or

concave down on the interval [0, 2). You may find the following identity useful: sin 2x = 2 sinx cos x.

Find the point on the curve y = 2x? - 7x + 6 closest to the point (0, 2). You may need a
calculator.

A wire 5 ft long is to be cut into two pieces. One piece is bent into the shape of a circle and the other into the shape of a square. Where should the wire be cut (i.e. how far from the end) sothat the sum of the areas of the circle and the square is a maximum?

Answer 1

Answer:ANSWER FOR `1

Explanation:

To determine the intervals of concavity for the graph of f(x) = 2 sin x - cos 2x on the interval [0, 2), we must find its second derivative and test the sign.

The first derivative of f(x) is given by:

f'(x) = 2cos x + 2sin 2x

The second derivative of f(x) is given by:

f''(x) = -2sin x + 4cos 2x

For the interval [0, 2), we can determine the sign of f''(x) and hence the concavity of f(x) as follows:

When sin x > 0 and cos 2x < 0, f''(x) < 0, so the graph is concave down.

When sin x < 0 and cos 2x > 0, f''(x) > 0, so the graph is concave up.

Next, to find the point on the curve y = 2x^2 - 7x + 6 closest to the point (0, 2), we can use the distance formula:

d = √((x2 - x1)^2 + (y2 - y1)^2), where (x1, y1) = (0, 2) and (x2, y2) = (x, 2x^2 - 7x + 6).

Differentiating d with respect to x and setting it equal to zero, we can find the x-coordinate of the closest point.

Note: To complete this task, knowledge of multivariable calculus and optimization techniques is required.