f(0)=1
f'(x)=3(2x-3)(x^2-3x+1)^2, f'(0)=-9
f''(x)=6(x^2-3x+1)^2+6(x^2-3x+1)(2x-3)^2, f''(0)=60
f'''(x)=12(x^2-3x+1)(2x-3)+6(2x-3)^3+24(x^2-3x+1)(2x-3)
f'''(0)=-270
f<4>(x)=12(2x-3)^2+24(x^2-3x+1)+36(2x-3)^2+24(2x-3)^2
+48(x^2-3x+1)=72(2x-3)^2+72(x^2-3x+1)
f<4>(0)=720
f<5>(x)=288(2x-3)+72(2x-3)
f<5>(0)=-1080
f<6>(x)=576+144=720
f<6>(0)=720
f<n>(x)=0,(当n>6)
所以f(x)=1+(-9)x+(60/2!)x^2+(-270/3!)x^3+(720/4!)x^4
+(-1080/5!)x^5+(720/6!)x^6
= 1 - 9x + 30x^2 - 45x^3 + 30x^4 - 9x^5 + x^6
追问谢啦
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这题怎么做?可以教我吗
那个f'''(x)后面为什么是24(x^2-3x+1)(2x-3),我怎么算也是12(……)
你好
在吗
能回答我的问题吗
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