作业帮大一高数微分中值定理求详解!
来源：学生作业帮 编辑：拍题作业网作业帮 分类：数学作业 时间：2024/04/29 18:58:14
大一高数微分中值定理求详解!
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(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(0)=720
f(x)=288(2x-3)+72(2x-3)
f(0)=-1080
f(x)=576+144=720
f(0)=720
f(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
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(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(0)=720
f(x)=288(2x-3)+72(2x-3)
f(0)=-1080
f(x)=576+144=720
f(0)=720
f(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