设函数f(x)=x²+2x，x∈R，求f(2)、f(-2)、f(a)、f(-a). 设函数f(x)=x²+|2x-a|(x∈R,a为常...
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设函数f(x)=x²+2x，x∈R，求f(2)、f(-2)、f(a)、f(-a).
f(2)=2²+2x2=4+4=8
f(-2)=（-2）²+2x（-2）=4-4=0
f（a）=a²+2a
f（-a）=（-a）²+2x（-a）=a²-2a
对于这个函数求解的时候，只需要把我们给定的定义域定义值，然后代入到公式当中求解，即可得出结果。
对于求函数值的问题，我们首先要了解函数值如何求解，其实就是将x的位置都换成那个常数再化简就行了，因为f(x)=x^2+2x，所以f(2)=2^2+2×2=8。f(－2)=(－2)^2+2×(－2)=0,f(a)=a^2+2a,f(－a)=(－a)^2+2×(－a)=a^2－2a
按照函数表达式，直接代进去一个个计算就OK了啊！
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绛旓細瑙ｇ瓟锛歞f(x)=f'(x)dx=x²dx
