Determine the points of inflection of the function.

f(x) = x + sin x (−2π ≤ x ≤ 2π)

Points of inflection are points on the curve where the curvature changes sign.

Since the sign of curvature is equal to the sign of the second derivative, so the points of inflection can be found to be where the second derivative changes sign.

For
f(x)=x+sin(x),
f'(x)=1+cos(x)
f"(x)=-sin(x)
Make a plot of -sin(x) and you should find that f"(x) changes sign at x=0,π and 2π.

The inflection points are therefore at __, __, and __ according to the definition.

y=100 find the first and second derivative

To determine the points of inflection of the function f(x) = x + sin(x) on the interval −2π ≤ x ≤ 2π, we need to find the second derivative and then identify the values of x where it changes sign.

Step 1: Find the first derivative of f(x):
f'(x) = 1 + cos(x)

Step 2: Find the second derivative of f(x) by differentiating the first derivative:
f''(x) = -sin(x)

Step 3: Set f''(x) = 0 and solve for x to find the critical points:
-sin(x) = 0
x = 0, π, 2π

Step 4: Check the sign of the second derivative in the intervals formed by the critical points:

For the interval −2π ≤ x < 0:
Choose a value x' such that −2π < x' < 0, for example x' = -π/2:
f''(-π/2) = -sin(-π/2) = -(-1) = 1
Since f''(x') > 0, the concavity is upward in this interval.

For the interval 0 < x < π:
Choose a value x' such that 0 < x' < π, for example x' = π/2:
f''(π/2) = -sin(π/2) = -1
Since f''(x') < 0, the concavity is downward in this interval.

For the interval π < x < 2π:
Choose a value x' such that π < x' < 2π, for example x' = 3π/2:
f''(3π/2) = -sin(3π/2) = -(-1) = 1
Since f''(x') > 0, the concavity is upward in this interval.

Step 5: The points of inflection are the x-values where the concavity changes, i.e., where the second derivative changes sign. From the above analysis, we can see that the concavity changes at x = 0 and x = π, so these are the points of inflection.

Therefore, the points of inflection for the function f(x) = x + sin(x) on the interval −2π ≤ x ≤ 2π are x = 0 and x = π.

To determine the points of inflection of a function, we need to find where the concavity changes. In other words, we want to find the x-values where the second derivative changes sign.

Let's find the first and second derivatives of the function f(x) = x + sin(x):

First derivative:
f'(x) = 1 + cos(x)

Second derivative:
f''(x) = -sin(x)

Now, we need to find where the second derivative changes sign. In this case, the second derivative f''(x) = -sin(x) is equal to zero when sin(x) = 0. This happens at x = 0, π, and -π.

To determine the sign changes, we can consider intervals:

Interval 1: (-2π, 0)
sin(x) is negative, so f''(x) is positive.

Interval 2: (0, π)
sin(x) is positive, so f''(x) is negative.

Interval 3: (π, 2π)
sin(x) is positive, so f''(x) is negative.

Therefore, the points of inflection for the given function f(x) = x + sin(x) are x = 0, π, and -π.