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Ilya [14]
3 years ago
8

This Test: 12 pts poss

Mathematics
1 answer:
Debora [2.8K]3 years ago
6 0

Answer:

Ni idea la verdad no entiendo

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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:

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Now:

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={ 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:

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