How do you find the inflection points for the function #f(x)=8x+3-2 sinx#?

Question

Charles Anctil · Accepted Answer

#f(x)=8x+3-2 sinx# #f'(x)=8-2 cosx# #f''(x)= 2 sinx# #f''(x) =2sinx = 0# at every integer multiple of #pi# and, we know the sine changes sign at every #x# intercept,  so every point:
#(n pi,0)# with integer #n# is an inflection point.

Charles Anctil · Answer

undefined To find the inflection points of the function $f(x) = 8x + 3 - 2 \sin(x)$, we need to find the points where the concavity changes. Inflection points occur where the second derivative changes sign or where the second derivative is equal to zero.

First, find the first derivative of $f(x)$:
$$f'(x) = 8 - 2 \cos(x)$$

Next, find the second derivative of $f(x)$:
$$f''(x) = 2 \sin(x)$$

To find where the concavity changes, set $f''(x) = 0$:
$$2 \sin(x) = 0$$
$$\sin(x) = 0$$

This happens at $x = 0, \pi, 2\pi, \ldots$.

To determine the concavity on either side of these points, test a value in the interval $(0, \pi)$, $(\pi, 2\pi)$, etc., in the second derivative. For example, test $x = \frac{\pi}{2}$ in $f''(x)$:
$$f''\left(\frac{\pi}{2}\right) = 2 \sin\left(\frac{\pi}{2}\right) = 2(1) = 2$$

Since $f''\left(\frac{\pi}{2}\right) > 0$, the function is concave up on the interval $(0, \pi)$.

Therefore, the inflection points occur at $x = 0, \pi, 2\pi, \ldots$, and the function is concave up on the intervals $(2n\pi, (2n + 1)\pi)$ and concave down on the intervals $((2n - 1)\pi, 2n\pi)$, where $n$ is an integer.