作业帮设函数f(x)在[a,b]上三阶可导,证明：存在一点e∈(a,b),使得f(b) = f(a) + 1/2 (b-a) [f'(a) + f'(b)] - 1/12 (b - a)^3 * f'''(e)
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![设函数f(x)在[a,b]上三阶可导,证明：存在一点e∈(a,b),使得f(b) = f(a) + 1/2 (b-a) [f'(a) + f'(b)] - 1/12 (b - a)^3 * f'''(e)](/uploads/image/z/5348436-60-6.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%E5%9C%A8%5Ba%2Cb%5D%E4%B8%8A%E4%B8%89%E9%98%B6%E5%8F%AF%E5%AF%BC%2C%E8%AF%81%E6%98%8E%EF%BC%9A%E5%AD%98%E5%9C%A8%E4%B8%80%E7%82%B9e%E2%88%88%28a%2Cb%29%2C%E4%BD%BF%E5%BE%97f%28b%29+%3D+f%28a%29+%2B+1%2F2+%28b-a%29+%5Bf%27%28a%29+%2B+f%27%28b%29%5D+-+1%2F12+%28b+-+a%29%5E3+%2A+f%27%27%27%28e%29)
设函数f(x)在[a,b]上三阶可导,证明：存在一点e∈(a,b),使得f(b) = f(a) + 1/2 (b-a) [f'(a) + f'(b)] - 1/12 (b - a)^3 * f'''(e)
设函数f(x)在[a,b]上三阶可导,证明：存在一点e∈(a,b),使得
f(b) = f(a) + 1/2 (b-a) [f'(a) + f'(b)] - 1/12 (b - a)^3 * f'''(e)
设函数f(x)在[a,b]上三阶可导,证明：存在一点e∈(a,b),使得f(b) = f(a) + 1/2 (b-a) [f'(a) + f'(b)] - 1/12 (b - a)^3 * f'''(e)
大部分就基于上楼的想法了,
f``(b)-f``(a)=(b-a)f```(e3)
f''(a)/2!((b-a)/2)² - f''(b)/2!((a-b)/2)²=-((b-a)/2)³f'''(e3)
f'''(e1)/3!((b-a)/2)³+f'''(e2)/3!((b-a)/2)³-((b-a)/2)³f'''(e3)=- f'''(e) ((b-a)/2)³/3
=(1/6+1/6-1)((b-a)/2)³ * f'''(e)=-1/12 (b - a)^3 * f'''(e)
f(x)=f(x0)+f'(x0)(x-x0)+f''(x0)/2! (x-x0)²+f'''(e1)/3! (x-x0)³
取x=(a+b)/2,x0=a,x0=b
f((a+b)/2)=f(a)+f'(a)((b-a)/2)+f''(a)/2! ((b-a)/2)²+f'''(e1)/3! ((b-a)/2)³
f((a+b)/2...
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f(x)=f(x0)+f'(x0)(x-x0)+f''(x0)/2! (x-x0)²+f'''(e1)/3! (x-x0)³
取x=(a+b)/2,x0=a,x0=b
f((a+b)/2)=f(a)+f'(a)((b-a)/2)+f''(a)/2! ((b-a)/2)²+f'''(e1)/3! ((b-a)/2)³
f((a+b)/2)=f(b)+f'(b)((a-b)/2)+f''(b)/2! ((a-b)/2)²+f'''(e2)/3! ((a-b)/2)³
相减，得
f(b)-f(a)=1/2 (b-a) [f'(a) + f'(b)]+f'''(e1)/3! ((b-a)/2)³-f'''(e2)/3! ((a-b)/2)³
f(b)=f(a)+1/2 (b-a) [f'(a) + f'(b)]+1/6 ((b-a)/2)³[f'''(e1)+f'''(e2)]
=f(a)+1/2 (b-a) [f'(a) + f'(b)]+1/6 ((b-a)/2)³[2f'''(e)]
=f(a)+1/2 (b-a) [f'(a) + f'(b)]+1/24(b-a)³f'''(e)
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