Question

Test yourself 2

15) if f''(x)=6x+6 and there is a stationary point at (0,3), find the equation of the curve

Answer 1

`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}`

Answer 2

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`