$f(x)$ = $\int ax^2+bx+c$  $dx$
= ${\frac{ax^3}{3}+\frac{bx^2}{2}+cx+d}$
$f'(x)$ = ${ax^2+bx+c}$
$f''(x)$ = ${2ax+b}$

$f(0)$ = $1$
$d$ = $1$
$f(1)$ = $1$
${\frac{a}{3}+\frac{b}{2}+1}$ = $0$
${2a+3b}$ = -$6$.......$(1)$
$f''(1)$ = $2$
${2a+b}$ = $2$.....$(2)$
$(1)$-$(2)$  ${2b = -8}$
${b = -4}$
${a = 3}$
$f'(0)$ = $1$
$c$= $1$
ดังนั้น
${f(x) = x^3-2x^2+x+1}$
${f'(x) = 3x^2-4x+1}$
${f'(x) = 0}$
${3x^2-4x+1}$ = $0$
$(3x-1)(x-1)$ = $0$
${x = \frac{1}{3}, 1}$
$f(\frac{1}{3})$ = ${3(\frac{1}{3})^3-4(\frac{1}{3})^2+\frac{1}{3}+1}$ = $\frac{31}{27}$
$f(1)$ = ${3(1)^3-4(1)^2+1+1}$ = $1$