一、求函数解析式【例1】设y=f(x)为三次函数,且图象关于原点对称,当x=1时,f(x)取得极小值-2,求f(x)的解析式.解:设f(x)=ax3+bx2+cx+d(a≠0),由于其关于原点对称,为奇函数.故b=d=0.所以f(x)=ax3+cx,由f′(x)=3ax2+c,且x=1时,f(x)有极小值-2得f′(1)=3a+c=0,f(1)=a+c=-2,解之,得 First, find the function analytical formula [Example 1] Let y = f (x) as a cubic function, and the image is symmetrical about the origin, when x = 1, f(x) obtains a minimum value of -2, find f(x) Analytic expression. Solution: Let f(x)=ax3+bx2+cx+d(a≠0), because it is symmetric about the origin, it is an odd function. So b=d=0. So f(x)=ax3+ Cx, from f′(x)=3ax2+c, and when x=1, f(x) has a minimum value of −2 to obtain f′(1)=3a+c=0, f(1)=a+c =-2, solution, get