Given: f (x) = sin2 2x; [0,π] a) Use a sign analysis to show the intervals where f(x) is increasing, and decreasing, and write the corresponding intervals using interval notation. B) Use a sign analysis to show the intervals where f(x) is concave up, and concave down, and write the corresponding intervals using interval notation. C) Locate the x-coordinate of all inflection points

A) Increasing from 0 to π/4

Decreasing from π/4 to π/2

B) f(x) is concave upwards on the interval, π/4 < x < (3/4)·π

f(x) is concave downwards on the interval, 0 < x < π/2

C) Where n = 0, 1, 2, ..., the inflection points are x = π/8 + n·π/2

Step-by-step explanation:

A) The given function is,  f(x) = sin²2·x, (0, π)

We note that at x = 0, f(0) = sin²(2×0) = 0

The maximum value of the sine function is sin(π/2), therefore, at the maximum value, we have;

2·x = π/2

x = π/4

At x = π/4, we have;

The maximum value, f(π/4) = sin²(2×π/4) = 1

At x = π/8, we have; f(π/8) = sin²(2×π/8) = 0.5

Using sign analysis, we have;

(f(π/4) - f(0))/(1 - 0) = (1 - 0)/(π/4 - 0) ≈ 1.27 (The slope is positive between 0 and π/4) and therefore, the function is increasing from 0 to π/4

Between π/4 and π/2, we have;

f(π/2) = sin²(2×π/2) = 0

Therefore, the slope, m = (0 - 1)/(π/2- π/4) ≈ -1.27 (The slope is negative between π/4 and π/2) and therefore, the function is decreasing from π/4 to π/2

B) The function is concave up when the slope is negative to the left of a point and positive to the right of the same point

Taking the point π/2, we have;

To the left, x = π/4, and the slope is (0 - 1)/(π/2- π/4) ≈ -1.27 is negative

To the right, x = (3/4)·π, f((3/4)·π) = sin²((3/4)·π) = 1, therefore;

The slope = (1 - 0)/((3/4)·π - π/2) ≈ 1.27, the slope is positive

Therefore, in the interval π/4 < x < (3/4)·π, the curve is concave upwards

The concave down where the slope to the left of a point is positive (the function is increasing) and the slope to the right is negative (the function is decreasing)

Therefore, the function is concave down in the interval, 0 < x < π/2

C) The inflection points are the turning points which are the points across which the slope changes sign

At the inflection point, we have;

f''(x) = d²(sin²2·x)/dx = 8·cos(4·x) = 0

∴ cos(4·x) = 0

4·x = arccos(0) = π/2

∴ The inflection points x = π/8 + n·π/2

i think it is 70

step-by-step explanation:

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.