题目内容

18.已知函数f(x)=$\left\{\begin{array}{l}{x+2,x≤-2}\\{{x}^{2},-2<x<2}\\{2x,x≥2}\end{array}\right.$,
(1)求f(-3),f[f(-3)].
(2)若f(a)=8,求a的值.

分析 (1)判断x的值所在的范围,代入分段函数求解即可;
(2)由f(a)=8可得$\left\{{\begin{array}{l}{a≤-2}\\{a+2=8}\end{array}}\right.$或$\left\{{\begin{array}{l}{-2<a<2}\\{{a^2}=8}\end{array}}\right.$或$\left\{{\begin{array}{l}{a≥2}\\{2a=8}\end{array}}\right.$,从而解得.

解答 解:(1)f(-3)=-3+2=-1,
f[f(-3)]=f(-1)=(-1)2=1;
(2)∵f(a)=8,
∴$\left\{{\begin{array}{l}{a≤-2}\\{a+2=8}\end{array}}\right.$或$\left\{{\begin{array}{l}{-2<a<2}\\{{a^2}=8}\end{array}}\right.$或$\left\{{\begin{array}{l}{a≥2}\\{2a=8}\end{array}}\right.$,
解得,a=4.

点评 本题考查了分段函数的一般解法及分类讨论的思想应用.

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