So, dy/dx=0 at the point (1, 4) - that is where x=1 and y=4.
\(\int { 12x-24dx } \\ \\ =\frac { 12{ x }^{ 2 } }{ 2 } -24x+C\\ \\ =6{ x }^{ 2 }-24x+C\)
\(\\ \\ \therefore \quad { f }^{ ' }\left( x \right) =6{ x }^{ 2 }-24x+C\)
But when x=1, f'(x)=0, therefore:
\(0=6-24+C\\ \\ 0=-18+C\\ \\ \therefore \quad C=18\)
\(\\ \\ \therefore \quad { f }^{ ' }\left( x \right) =6{ x }^{ 2 }-24x+18\)
Now:
\(\int { 6{ x }^{ 2 } } -24x+18dx\\ \\ =\frac { 6{ x }^{ 3 } }{ 3 } -\frac { 24{ x }^{ 2 } }{ 2 } +18x+C\)
\(=2{ x }^{ 3 }-12{ x }^{ 2 }+18x+C\\ \\ \therefore \quad f\left( x \right) =2{ x }^{ 3 }-12{ x }^{ 2 }+18x+C\)
Now when x=1, y=4:
\(4=2-12+18+C\\ \\ 4=8+C\\ \\ C=4-8\\ \\ C=-4\)
\(\\ \\ \therefore \quad f\left( x \right) =2{ x }^{ 3 }-12{ x }^{ 2 }+18x-4\)
Now when x=2,
\(f\left( x \right) =2\cdot { 2 }^{ 3 }-12\cdot { 2 }^{ 2 }+18\cdot 2-4\\ \\ =16-48+36-4\\ \\ =0\)
So when x=2, y=0.
