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Alenkinab [10]
3 years ago
10

Test yourself 2

Mathematics
2 answers:
mafiozo [28]3 years ago
7 0
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
antoniya [11.8K]3 years ago
7 0
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}

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