写出 $\tan^{-1} x$ 的主值范围。 下方显示了在 $0 \le x \le \tfrac{3\pi}{2}$ 弧度范围内的曲线 $y = a + b \cos c x$。 该曲线的极大值点为 $\left(\tfrac{\pi}{2},7\right)$ 和 $\left(\tfrac{3\pi}{2},7\right)$。 该曲线的极小值点为 $(0,1)$ 和 $(\pi,1).$ 下方显示了在 $0 \le x \le \tfrac{3\pi}{2}$ 弧度范围内的曲线 $y = a + b \cos c x$。 该曲线的极大值点为 $\left(\tfrac{\pi}{2},7\right)$ 和 $\left(\tfrac{3\pi}{2},7\right)$。 该曲线的极小值点为 $(0,1)$ 和 $(\pi,1).$ 求 $a$、$b$ 和 $c$ 的值。 求 $a$、$b$ 和 $c$ 的值。 在同一坐标系中，画出一条直线以确定方程 $a\pi + b\pi \cos c x = 5\pi - 2x$ 在 $0 \le x \le \tfrac{3\pi}{2}$ 范围内的解的个数。

1. State the range of principal values for the inverse tangent function

The principal values of $\tan^{-1} x$ lie strictly between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$.

$-\frac{\pi}{2} < \tan^{-1} x < \frac{\pi}{2}$

1. Find the value of $a$ (the vertical shift).

The maximum value is $7$ and the minimum value is $1$.
$a = \frac{\text{max} + \text{min}}{2} = \frac{7 + 1}{2} = 4$

2. Find the value of $b$ (related to amplitude).

The amplitude is $\frac{\text{max} - \text{min}}{2} = \frac{7 - 1}{2} = 3$.
Since the curve has a minimum at $x = 0$ instead of a maximum, the cosine curve is reflected about its principal axis.
Therefore, $b = -3$.

3. Find the value of $c$ (related to the period).

The curve completes one full cycle between the minimum points at $x = 0$ and $x = \pi$.
Period $= \pi - 0 = \pi$
We know that Period $= \frac{2\pi}{c}$.
$\frac{2\pi}{c} = \pi \implies c = 2$

1. Find the values of $a$, $b$, and $c$ from the given graph properties

From the graph, the maximum value is $7$ and the minimum is $1$.
Amplitude $= \frac{7 - 1}{2} = 3$.
Since the curve starts at a minimum at $x=0$, $b = -3$.
Vertical shift $a = \frac{7 + 1}{2} = 4$.
The period is $\pi$ (from minimum at $0$ to minimum at $\pi$), so $c = \frac{2\pi}{\pi} = 2$.
Thus, the curve is $y = 4 - 3\cos 2x$.

2. Rearrange the given equation to match the curve's equation on one side

The given equation is $a\pi + b\pi \cos cx = 5\pi - 2x$.
Substitute $a=4$, $b=-3$, and $c=2$:
$4\pi - 3\pi \cos 2x = 5\pi - 2x$
Divide the entire equation by $\pi$:
$4 - 3\cos 2x = 5 - \frac{2}{\pi}x$
The left side is exactly the equation of the curve $y = 4 - 3\cos 2x$.
Therefore, we need to draw the straight line $y = 5 - \frac{2}{\pi}x$.

3. Determine the number of intersection points between the curve and the line

Plot the line $y = 5 - \frac{2}{\pi}x$ on the same axes:

• At $x = 0$, $y = 5$
• At $x = \frac{\pi}{2}$, $y = 5 - 1 = 4$
• At $x = \pi$, $y = 5 - 2 = 3$
• At $x = \frac{3\pi}{2}$, $y = 5 - 3 = 2$

Tracing the intersections:

1. Between $x=0$ and $x=\frac{\pi}{2}$, the curve goes from $1$ to $7$, while the line goes from $5$ to $4$. (1 intersection)
2. Between $x=\frac{\pi}{2}$ and $x=\pi$, the curve goes from $7$ to $1$, while the line goes from $4$ to $3$. (1 intersection)
3. Between $x=\pi$ and $x=\frac{3\pi}{2}$, the curve goes from $1$ to $7$, while the line goes from $3$ to $2$. (1 intersection)

Total number of intersections = 3.