Please help with derivatives and functions

Hello,

I`m a total noob in math and barely remember anything from high school past one variable equations.

If someone can help me with several issues here, I`d be most appreciated.

It is given that the second derivative of a function is g"(x) = 4

and that g`(0) = 0 and g(0) = 4.

What`s the function g(x) ?

Any help is most appreciated.

Thanks

Re: Please help with derivatives and functions

We haveg''(x) = 4, g'(0) = 0 and g(0) = 4

Start with the function g''(x) = 4 and find its primitive functiong''(x) = 4 => g'(x) = 4x+C

Since g'(0) = 4 we have g'(x) = 4x+C => 4*0+C = 0 <=> C = 0

Finding the primitive function of f'(x) gives usg'(x) = 4x => g(x) = 2x^2+D

Since g(0) = 4 we haveg(x) = 2x^2+D => 4 = 2*0^2+D <=> D = 4

Thus, g(x) = 2x^2+4

Re: Please help with derivatives and functions

$\displaystyle g''(x) = 4$

antiderivative ...

$\displaystyle g'(x) = 4x + C$

$\displaystyle g'(0) = 0 \implies C = 0$

$\displaystyle g'(x) = 4x$

antiderivative ...

$\displaystyle g(x) = 2x^2 + C$

$\displaystyle g(0) = 4 \implies C = 4$

$\displaystyle g(x) = 2x^2 + 4$

next time, post calculus questions in the calculus forum ... the lobby is just a place for introductions.

Re: Please help with derivatives and functions

Quote:

Originally Posted by

**fkf** We have

g''(x) = 4, g'(0) = 4 and g(0) = 4

The problem gave g'(0)= 0.

Quote:

Start with the function g''(x) = 4 and find its primitive function

g''(x) = 4 => g'(x) = 4x+C

Since g'(0) = 4 we have

g'(x) = 4x+C => 4*0+C = 4 <=> C = 4

Finding the primitive function of f'(x) gives us

g'(x) = 4x+4 => g(x) = 2x^2+4x+D

Since g(0) = 4 we have

g(x) = 2x^2+4x+D => 4 = 2*0^2+4*0+D <=> D = 4

Thus, g(x) = 2x^2+4x+4

Re: Please help with derivatives and functions

Oh, my bad. Sorry for that.

Re: Please help with derivatives and functions

Thanks guys.

It was very helpful.

Next time I`ll post in the appropriate forum.