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Answer 1

\(f'(x)=2x^3\), find f. Given the tangent 2x+y=0
First the general anti-derivative: \(F(x)=\frac{1}{2}x^4+c\)
Got that \(y=-2x\), slope \(m=-2\), or f'(x)=-2\) right?

So \(-2=2x^3\) and solving for x.. \(-1=x\).
Plug that into the tangent \(2(-1)+y=0\) and get \(y=2\), giving the points \((-1, 2)\).

Answer 2

Finishing by plugging it back into \(F(x)=\frac{1}{2}x^4\) so.. \(2=\frac{1}{2}(-1)^4+C\)
So \(C=1.5\)?

Answer 3

seems correct to me :)

Answer 4

Alright, thank you hartnn :D