Question

question 5
find the inflection points for the function given.
f(x)=7x + 2 - sinx, 0<x<3π
(π,7π + 2), (2π,14π)
(π,7π), (2π,14π + 2)
(π,7π), (2π,14π)
(π,2), (2π,14π + 2)
(π,7π + 2), (2π,14π + 2)

Explanation:

Step1: Find the first - derivative

The derivative of $y = 7x+2-\sin x$ is $y'=\frac{d}{dx}(7x)+\frac{d}{dx}(2)-\frac{d}{dx}(\sin x)=7 - \cos x$.

Step2: Find the second - derivative

Differentiate $y' = 7-\cos x$ with respect to $x$. The derivative of $y'$ is $y''=\frac{d}{dx}(7)-\frac{d}{dx}(\cos x)=\sin x$.

Step3: Solve for inflection points

Inflection points occur where $y'' = 0$ and $y''$ changes sign. Set $y''=\sin x = 0$ in the interval $0\lt x\lt3\pi$. The solutions of $\sin x = 0$ in this interval are $x=\pi$ and $x = 2\pi$.
To find the $y$ - values, substitute $x=\pi$ and $x = 2\pi$ into the original function $y = 7x+2-\sin x$.
When $x=\pi$, $y=7\pi+2-\sin(\pi)=7\pi + 2$.
When $x = 2\pi$, $y=7\times(2\pi)+2-\sin(2\pi)=14\pi+2$.

Answer:

$(\pi,7\pi + 2),(2\pi,14\pi+2)$