函数 $\text{g}$ 定义为 $\text{g} : x \to \sin x + x$ ，其中 $x \in \mathbb{R}$ ， $-\pi \le x \le \pi$ 。

(i)
证明对于 $-\pi < x < \pi$ ，有 $\text{g}'(x) > 0$ 。由此或用其他方法，证明 $\text{g}$ 存在反函数。
(ii)
在同一坐标系中草拟 $y = \text{g}(x)$ 和 $y = \text{g}^{-1}(x)$ 的图像。

存在性条件反函数图像导数的几何意义函数概念 English

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完整解法

2(i)

1. Find the first derivative of $\text{g}(x)$.

$\text{g}'(x) = \frac{\text{d}}{\text{d}x}(\sin x + x) = \cos x + 1$

2. Show that the derivative is strictly positive on the given open interval.

For $-\pi < x < \pi$ , the range of $\cos x$ is $-1 < \cos x \le 1$ .
Since $\cos x > -1$ , it follows that $\cos x + 1 > 0$ .
Hence, $\text{g}'(x) > 0$ for $-\pi < x < \pi$ .

3. Deduce that the function has an inverse.

Since $\text{g}'(x) > 0$ for the interior of its domain, $\text{g}$ is a strictly increasing function.
A strictly increasing function is a one-to-one function (it passes the horizontal line test).
Therefore, the inverse function $\text{g}^{-1}$ exists.

答案

Shown.

2(ii)

1. Identify key coordinates, relative position to $y=x$, and endpoint tangents of $y = \text{g}(x)$.

$\text{g}(-\pi) = -\pi, \quad \text{g}(0) = 0, \quad \text{g}(\pi) = \pi$
Since $\text{g}'(\pm\pi) = \cos(\pm\pi) + 1 = 0$ , the graph of $\text{g}$ has horizontal tangents at the endpoints $(-\pi, -\pi)$ and $(\pi, \pi)$ .
Furthermore, for $0 < x < \pi$ , $\sin x > 0 \implies \text{g}(x) > x$ , so the curve lies above the line $y=x$ . Conversely, for $-\pi < x < 0$ , $\sin x < 0 \implies \text{g}(x) < x$ , meaning the curve lies below $y=x$ .

2. Deduce the corresponding properties for $y = \text{g}^{-1}(x)$ using the reflection property of inverse functions.

The graph of $y = \text{g}^{-1}(x)$ is the reflection of $y = \text{g}(x)$ in the line $y = x$ .
Therefore, it will share the intersection points $(-\pi, -\pi)$ , $(0,0)$ , and $(\pi, \pi)$ .
The horizontal tangents of $\text{g}(x)$ become vertical tangents for $\text{g}^{-1}(x)$ at the endpoints.
For $0 < x < \pi$ , $\text{g}^{-1}(x)$ lies below $y=x$ , and for $-\pi < x < 0$ , it lies above $y=x$ .

3. Produce the combined sketch.

(Student draws the axes, sketching the line $y=x$ as a guide, and carefully draws $y = \text{g}(x)$ and $y = \text{g}^{-1}(x)$ exhibiting the correct intersection points, relative positions, and tangent behaviors at the extremities.)

答案

A sketch containing both curves on the interval $[-\pi, \pi]$ . They intersect at $(-\pi, -\pi)$ , $(0, 0)$ , and $(\pi, \pi)$ .

$y = \text{g}(x)$ has horizontal tangents at the endpoints and lies above $y=x$ for $0 < x < \pi$ .
$y = \text{g}^{-1}(x)$ has vertical tangents at the endpoints and lies below $y=x$ for $0 < x < \pi$ .

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函数 g\text{g}g 定义为 g:x→sin⁡x+x\text{g} : x \to \sin x + xg:x→sinx+x，其中 x∈Rx \in \mathbb{R}x∈R，−π≤x≤π-\pi \le x \le \pi−π≤x≤π。

完整解法

2(i)

2(ii)

同一知识点的其他题

函数 $\text{g}$ 定义为 $\text{g} : x \to \sin x + x$ ，其中 $x \in \mathbb{R}$ ， $-\pi \le x \le \pi$ 。