题目内容
7.已知函数$f(x)=\left\{{\begin{array}{l}{{x^2}+1}\\{{2^x}}\end{array}}\right.\begin{array}{l}{(x≤0)}\\{(x>0)}\end{array}$,则满足f(x)=4的x的取值是2或$-\sqrt{3}$.分析 由已知条件结合分段函数的性质得当x≤0时,x2+1=4;当x>0时,2x=4.由此能求出满足f(x)=4的x的取值.
解答 解:∵函数$f(x)=\left\{{\begin{array}{l}{{x^2}+1}\\{{2^x}}\end{array}}\right.\begin{array}{l}{(x≤0)}\\{(x>0)}\end{array}$,满足f(x)=4,
∴当x≤0时,x2+1=4,解得x=$\sqrt{3}$(舍)或x=-$\sqrt{3}$;
当x>0时,2x=4,解得x=2.
∴满足f(x)=4的x的取值是2或$-\sqrt{3}$.
故答案为:2或$-\sqrt{3}$.
点评 本题考查函数值的求法,是基础题,解题时要认真审题,注意分段函数的性质的合理运用.
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