## Task B-3.3.

The function is given by $f(x)=A \sin \left(B x+\frac{7 \pi}{10}\right)+D$, where $A, B$, and $D$ are positive real numbers. The point $M\left(\frac{2 \pi}{3},-8\right)$ is the only minimum point, and the point $N\left(\frac{3 \pi}{2}, 12\right)$ is the only maximum point of the function $f$ in the interval $\left[\frac{2 \pi}{3}, \frac{3 \pi}{2}\right]$. Determine $f\left(\frac{203 \pi}{12}\right)$.
## Solution.

Let's plot points $M$ and $N$ in the coordinate system and sketch the graph of the function $f$.

![](https://cdn.mathpix.com/cropped/2024_05_30_d455efeee432fadf0574g-12.jpg?height=648&width=505&top_left_y=1935&top_left_x=707)

First, let's determine the coefficients $A$ and $D$.

The coefficient $D$ is the vertical shift of the sine function or calculated as:

$$
D=\frac{\text { "maximum value" }+ \text { "minimum value" }}{2}=\frac{12+(-8)}{2}=2
$$

The coefficient $A$ is the amplitude of the given function or calculated as:

$$
A=\frac{\text { "maximum value" }- \text { "minimum value" }}{2}=\frac{12-(-8)}{2}=10
$$

For the fundamental period $T$ of the function $f$, we have $\frac{T}{2}=\frac{3 \pi}{2}-\frac{2 \pi}{3}=\frac{5 \pi}{6}$, hence $T=\frac{5 \pi}{3}$.

Now we can determine the coefficient $B: B=\frac{2 \pi}{T}=\frac{2 \pi}{\frac{5 \pi}{3}}=\frac{6}{5}$.

Therefore, the given function is $f(x)=10 \sin \left(\frac{6}{5} x+\frac{7 \pi}{10}\right)+2$.

Finally, we need to determine $f\left(\frac{203 \pi}{12}\right)$.

$f\left(\frac{203 \pi}{12}\right)=10 \sin \left(\frac{6}{5} \cdot \frac{203 \pi}{12}+\frac{7 \pi}{10}\right)=10 \sin (21 \pi)+2=10 \cdot 0+2=2$.

Note: A student can achieve full points in the problem even if they have not sketched the graph of the function. A student can earn points for accurately determining the coefficients $A$ and $D$ if they read them from the sketch.