Answer:
[tex]f(x)=x^3-x^2+2x-1[/tex]
Step-by-step explanation:
Since [tex]f''(x)=6x-2[/tex], then [tex]f'(x)=3x^2-2x+C[/tex], thus:
[tex]f'(x)=3x^2-2x+C\\\\f'(1)=3(1)^2-2(1)+C\\\\3=3-2+C\\\\3=1+C\\\\2=C[/tex]
Therefore, since [tex]f'(x)=3x^2-2x+2[/tex], then [tex]f(x)=x^3-x^2+2x+C[/tex], thus:
[tex]f(x)=x^3-x^2+2x+C\\\\f(2)=2^3-2^2+2(2)+C\\\\7=8-4+4+C\\\\7=8+C\\\\-1=C[/tex]
In conclusion, [tex]f(x)=x^3-x^2+2x-1[/tex]
