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test yourself 2
15) if f''(x)=6x+6 and there is a stationary point at (0,3), find the equation of the curve

Sagot :

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

[tex]\int { 6x+6dx } \\ \\ =\frac { 6{ x }^{ 2 } }{ 2 } +6x+C\\ \\ =3{ x }^{ 2 }+6x+C[/tex]

[tex]\\ \\ \therefore \quad { f }^{ ' }\left( x \right) =3{ x }^{ 2 }+6x+C[/tex]

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

Therefore:

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

Now:

[tex]\int { 3{ x }^{ 2 } } +6xdx\\ \\ =\frac { 3{ x }^{ 3 } }{ 3 } +\frac { 6x^{ 2 } }{ 2 } +C[/tex]

[tex]={ x }^{ 3 }+3{ x }^{ 2 }+C\\ \\ \therefore \quad f\left( x \right) ={ x }^{ 3 }+3{ x }^{ 2 }+C[/tex]

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

[tex]3=C\\ \\ \therefore \quad f\left( x \right) ={ x }^{ 3 }+3{ x }^{ 2 }+3[/tex]
konrad509
[tex]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} [/tex]
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