分段函數的複合函數要怎麼求（1）

05-09

聲明：本文為原創文章，首發於微分公眾號「湖心亭記」

相關鏈接分段函數的複合函數要怎麼求（2）

求分段函數的複合函數，這是考研高數中的一個重要考點。專升本的高數不考這個。因此專升本考試的考生可略過。今天我們就來詳細說一下這樣的題目應該怎麼做。

有的老師會給大家提供用幾何方法（數形結合方法來做）。但是我個人是不太喜歡畫圖的。一個是因為分段函數畫圖有時候容易畫圖，一個是因為畫圖的方法不好形成具體的做題步驟。所以我介紹的是用代數的方法來做的。

首先題目類型是：f(x)和g(x)是兩個分段函數f[g(x)]

我們的解題方法是：依據f(x)中的x的範圍，求出g(x)在該範圍下的表達式和對應自變數的範圍，然後回代替換即可。

好了，廢話不多說，我們先看一個例子。

例1 設 $[fleft( x ight) = left{ egin{array}{l} {e^x}{kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} x < 1\ {x^2} - 1{kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} x ge 1 end{array} ight.]$ ， $[gleft( x ight) = left{ egin{array}{l} x + 2{kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} x < 0\ {x^2} - 1{kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} x ge 0 end{array} ight.]$ ，求 $[fleft[ {gleft( x ight)} ight]]$

解析：（1）我們先寫出 $[fleft[ {gleft( x ight)} ight]]$ 的表達式。這種寫法要記住，形成格式。如下

$[fleft[ {gleft( x ight)} ight] = left{ egin{array}{l} {e^{gleft( x ight)}}{kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} gleft( x ight) < 1\ {left[ {gleft( x ight)} ight]^2} - 1{kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} gleft( x ight) ge 1 end{array} ight.]$

（2）下面我的目標就是要把上式中的 $[gleft( x ight)]$ 替換成具體的關於x的表達式，後面的範圍也替換成具體的x的範圍。所以我們的思路自然而然的是要尋找 $[gleft( x ight) < 1]$ 和 $[gleft( x ight) ge 1]$ 下的具體的x的表達式和範圍了。顯然要分類討論。

如下：

當 $[gleft( x ight) < 1]$ 時

① 當x<0

$[gleft( x ight) = x + 2 < 1{kern 1pt} {kern 1pt} {kern 1pt} Rightarrow {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} x < - 1]$

故此時 $[x < - 1]$ ，且 $[gleft( x ight) = x + 2]$

② 當 $[x ge 0]$

$[gleft( x ight) = {x^2} - 1 < 1 Rightarrow - sqrt 2 < x < sqrt 2 ]$

故 $[0 le x < sqrt 2 ]$ ，且 $[gleft( x ight) = {x^2} - 1]$

當 $[gleft( x ight) ge 1]$ 時

① 當x<0

$[gleft( x ight) = x + 2 ge 1{kern 1pt} {kern 1pt} {kern 1pt} Rightarrow {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} x ge - 1]$

故此時 $[ - 1 le x < 0]$ ，且 $[gleft( x ight) = x + 2]$

② 當 $[x ge 0]$

$[gleft( x ight) = {x^2} - 1 ge 1 Rightarrow x ge sqrt 2 ]$ 或 $[x le - sqrt 2 ]$

故 $[x ge sqrt 2 ]$ ，且 $[gleft( x ight) = {x^2} - 1]$

（3）好了，具體的範圍和表達式都求出來了。我們來做個總結如下

因此有 $[gleft( x ight) < 1]$ 時 $[left{ egin{array}{l} gleft( x ight) = x{ m{ + }}2{kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} x < - 1\ gleft( x ight) = {x^2} - 1{kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} 0 le x < sqrt 2 end{array} ight.]$

$[gleft( x ight) ge 1]$ 時 $[left{ egin{array}{l} gleft( x ight) = x{ m{ + }}2{kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} - 1 le x < 0\ gleft( x ight) = {x^2} - 1{kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} x ge sqrt 2 end{array} ight.]$

（4）按照我們的總結進行回代到（1）中的式子，如下

因此有 $[fleft[ {gleft( x ight)} ight] = left{ egin{array}{l} {e^{x + 2}}{kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} x < - 1\ {e^{{x^2} - 1}}{kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} 0 le x < sqrt 2 \ {left( {x + 2} ight)^2} - 1{kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} - 1 le x < 0\ {left( {{x^2} - 1} ight)^2} - 1{kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} x ge sqrt 2 end{array} ight.]$

整理後如下

$[fleft[ {gleft( x ight)} ight] = left{ egin{array}{l} {e^{x + 2}}{kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} x < - 1\ {left( {x + 2} ight)^2} - 1{kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} - 1 le x < 0\ {e^{{x^2} - 1}}{kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} 0 le x < sqrt 2 \ {left( {{x^2} - 1} ight)^2} - 1{kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} x ge sqrt 2 end{array} ight.]$

===================================

所以大家可以看到，嚴格遵循以上的四個步驟，邏輯嚴密，而且絕不會出錯。是不是顯得題目其實也簡單許多啊。

下面再舉一道例題，我就不詳細解出每一步的思路意圖了。直接給步驟答案，讓大家再體會一下分段函數求複合函數的解法。如下。