INTEGRATION十分難!!

d^2y/dx^2=6x-4

dy/dx=3x^2-4x+C

sub x=1,dy/dx=0

3-4+C=0
C=1

dy/dx=3x^2-4x+1=(x-1)(3x-1)

The equation of the curve is

x^3-2x^2+x+K where K is a constant

sub x=1, y=7

1-2+1+K=7

K=7

So x^3-2x^2+x+7

Let dy/dx=0,then x=1 or x=1/3

So the maximun point of y is (1/3)^3-2(1/3)^2+(1/3)+7

=1/27-1/18+1/3+7

d²y/dx²=6x-4
dy/dx=3x²-4x+C, where C is a constant
When x=1, y has a local minimum
∴3(1)²-4(1)+C=0
C=1
∴dy/dx=3x²-4x+1
∴y=x³-2x²+x+K, where K is a constant
When x=1, y=7
∴7=1-2+1+K
K=7
∴y=x³-2x²+x+7

dy/dx=0, 3x²-4x+1=0
(3x-1)(x-1)=0
x=1 or 1/3
d²y/dx²=6(1/3)-4=-2 < 0
∴x=1/3 is a max. pt.
∴Max. pt. of y=1/27-2/9+1/3+7=193/27