2024年高考数学上海2 |
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2024-08-28 22:56:07 |
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(4分)已知$f(x)=\left\{\begin{array}{l}{\sqrt{x},x > 0}\\ {1,x\leqslant 0}\end{array}\right.$,则$f(3)=$____.
答案:$\sqrt{3}$. 分析:根据已知条件,将$x=3$代入函数解析式,即可求解. 解:$f(x)=\left\{\begin{array}{l}{\sqrt{x},x > 0}\\ {1,x\leqslant 0}\end{array}\right.$, 则$f(3)=\sqrt{3}$. 故答案为:$\sqrt{3}$. 点评:本题主要考查函数的值,属于基础题.
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