What are the points of inflection of #f(x)=8x + 2xsinx
# on # x in [0, 2pi]#?

Question

What are the points of inflection of #f(x)=8x + 2xsinx
  # on # x in [0, 2pi]#?

Emilia Aguilar · Accepted Answer

#x_1 approx 1.07687398631180365861#
#x_2 approx 3.6435971674254006414# #f'(x)=8+2sin(x)+2xcos(x)# #f''(x)=4cos(x)-2xsin(x)# #f'''(x)=-6sin(x)-2xcos(x)# By a numerical method we get the Solutions above. and #f'''(x_1)ne 0# and #f'''(x_2) ne 0#

Ezra Adams · Answer

undefined To find the points of inflection of $ f(x) = 8x + 2x \sin(x) $ on $ x $ in $[0, 2\pi]$, we need to find where the concavity of the function changes. Points of inflection occur where the second derivative changes sign or where the second derivative is equal to zero.

The first derivative of $ f(x) $ is:
$$ f'(x) = 8 + 2 \sin(x) + 2x \cos(x) $$

The second derivative is:
$$ f''(x) = 2 \cos(x) - 2x \sin(x) + 2 \cos(x) - 2x \sin(x) $$

Simplify this to get:
$$ f''(x) = 4 \cos(x) - 2x \sin(x) $$

Set $ f''(x) $ equal to zero and solve for $ x $ to find critical points:
$$ 4 \cos(x) - 2x \sin(x) = 0 $$

There are two types of solutions: those where $ \cos(x) = 0 $ and those where $ \sin(x) = \frac{2}{x} $.

For $ \cos(x) = 0 $, we get $ x = \frac{\pi}{2}, \frac{3\pi}{2} $.

For $ \sin(x) = \frac{2}{x} $, we have to solve numerically for the other points.

Next, we analyze the sign of $ f''(x) $ in intervals between these critical points to determine concavity:

1. For $ x \in (0, \frac{\pi}{2}) $, $ f''(x) > 0 $, so the function is concave up.
2. For $ x \in (\frac{\pi}{2}, \frac{3\pi}{2}) $, $ f''(x) < 0 $, so the function is concave down.
3. For $ x \in (\frac{3\pi}{2}, 2\pi) $, $ f''(x) > 0 $, so the function is concave up.

Thus, the points of inflection are at $ x = \frac{\pi}{2} $ and $ x = \frac{3\pi}{2} $.

What are the points of inflection of #f(x)=8x + 2xsinx # on # x in [0, 2pi]#?