Question

Test yourself 2

Answer 1

Ok, so dy/dx=0 at the point (0,3) that is where x=0 and y=3.

$\int { 6x+6dx } \\ \\ =\frac { 6{ x }^{ 2 } }{ 2 } +6x+C\\ \\ =3{ x }^{ 2 }+6x+C$

$\\ \\ \therefore \quad { f }^{ ' }\left( x \right) =3{ x }^{ 2 }+6x+C$

Now, f'(x)=0 when x=0.

Therefore:

$0=C\\ \\ \therefore \quad { f }^{ ' }\left( x \right) =3{ x }^{ 2 }+6x$

Now:

$\int { 3{ x }^{ 2 } } +6xdx\\ \\ =\frac { 3{ x }^{ 3 } }{ 3 } +\frac { 6x^{ 2 } }{ 2 } +C$

$={ x }^{ 3 }+3{ x }^{ 2 }+C\\ \\ \therefore \quad f\left( x \right) ={ x }^{ 3 }+3{ x }^{ 2 }+C$

But when x=0, y=3, therefore:

$3=C\\ \\ \therefore \quad f\left( x \right) ={ x }^{ 3 }+3{ x }^{ 2 }+3$

Answer 2

$f''(x)=6x+6\\ f'(x)=\int 6x+6\, dx\\ f'(x)=3x^2+6x+C\\\\ 0=3\cdot0^2+6\cdot0+C\\ 0=C\\ f'(x)=3x^2+6x\\\\ f(x)=\int 3x^2+6x\, dx\\ f(x)=x^3+3x^2+C\\\\ 3=0^3+3\cdot0^2+C\\ 3=C\\ \\\boxed{f(x)=x^3+3x^2+3}$