Question

设f（x）在[0,1]上二阶导数连续,f（0）=f（1）=0,并且当x属于（0,1）时,|f‘‘（x）|小于等于A,求证

Answer 1

f(0)=f(x)-f'(x)x+f''(ξ）x^2 /2

f(1)=f(x)-f'(x)(x-1)+f''(η）(x-1)^2 /2

两式相减,f'(x)=[f''(ξ)*x^2 - f''(η)*(x-1)^2] /2

|f'(x)|