分段函數的複合函數要怎麼求（2）

05-16

因為知乎字數限制，發不出去，所以分成兩篇來發

分段函數的複合函數要怎麼求（1）

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

解：令 $[gleft( x ight) = fleft( x ight) = left{ egin{array}{l} ln sqrt x {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} x ge 1\ 2x - 1{kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} x < 1 end{array} ight.]$ ，則等價於求 $[fleft[ {gleft( x ight)} ight]]$

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

（2）當 $[gleft( x ight) ge 1]$

① $[x ge 1]$

$[gleft( x ight) = ln sqrt x ge 1 Rightarrow x ge {e^2}]$

故 $[x ge {e^2}]$

② $[x < 1]$

$[gleft( x ight) = 2x - 1 ge 1 Rightarrow x ge 1]$ ，空集

當 $[gleft( x ight) < 1]$

① $[x ge 1]$

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

故 $[1 le x < {e^2}]$

② $[x < 1]$ $[gleft( x ight) = 2x - 1 < 1 Rightarrow x < 1]$

故 $[x < 1]$

（3）因此有

$[gleft( x ight) ge 1]$ 時 $[gleft( x ight) = ln sqrt x {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} x ge {e^2}]$

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

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