Question

1. consider the function $y = \sin(5x)$ on the interval $0 < x < \frac{2\pi}{5}$

a) determine where the graph of the function is concave upward and where it is concave downward on the interval $0 < x < \frac{2\pi}{5}$
b) find all inflection points of the function on the interval $0 < x < \frac{2\pi}{5}$.

Explanation:

Step1: Find the first - derivative

Using the chain - rule, if $y = \sin(5x)$, then $y^\prime=\frac{d}{dx}(\sin(5x)) = 5\cos(5x)$.

Step2: Find the second - derivative

Differentiate $y^\prime = 5\cos(5x)$ with respect to $x$. Using the chain - rule again, $y^{\prime\prime}=\frac{d}{dx}(5\cos(5x))=- 25\sin(5x)$.

Step3: Determine concavity

The function is concave upward when $y^{\prime\prime}>0$ and concave downward when $y^{\prime\prime}<0$.
Set $y^{\prime\prime}=-25\sin(5x)>0$. Then $\sin(5x)<0$.
For $0 < x<\frac{2\pi}{5}$, we have $0 < 5x<2\pi$. Solving $\sin(5x)<0$ for $0 < 5x<2\pi$ gives $\pi<5x<2\pi$, or $\frac{\pi}{5}Set $y^{\prime\prime}=-25\sin(5x)<0$. Then $\sin(5x)>0$. Solving $\sin(5x)>0$ for $0 < 5x<2\pi$ gives $0 < 5x<\pi$, or $0 < x<\frac{\pi}{5}$. So the function is concave downward on the interval $(0,\frac{\pi}{5})$.

Step4: Find inflection points

Inflection points occur where $y^{\prime\prime} = 0$ and $y^{\prime\prime}$ changes sign.
Set $y^{\prime\prime}=-25\sin(5x)=0$. Then $\sin(5x)=0$. For $0 < x<\frac{2\pi}{5}$, when $5x=\pi$, $x = \frac{\pi}{5}$.
We can check the sign of $y^{\prime\prime}$ on either side of $x=\frac{\pi}{5}$. For $x\in(0,\frac{\pi}{5})$, $y^{\prime\prime}<0$ and for $x\in(\frac{\pi}{5},\frac{2\pi}{5})$, $y^{\prime\prime}>0$. So $x = \frac{\pi}{5}$ is an inflection point. When $x=\frac{\pi}{5}$, $y=\sin(5\times\frac{\pi}{5})=\sin(\pi)=0$. The inflection point is $(\frac{\pi}{5},0)$.

Answer:

a) Concave downward on $(0,\frac{\pi}{5})$, concave upward on $(\frac{\pi}{5},\frac{2\pi}{5})$.
b) Inflection point is $(\frac{\pi}{5},0)$.