# Please help with derivatives and functions

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• Nov 5th 2012, 06:37 AM
ZKR
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
• Nov 5th 2012, 06:46 AM
fkf
Re: Please help with derivatives and functions
We have
g''(x) = 4, g'(0) = 0 and g(0) = 4

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 = 0 <=> C = 0

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

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

Thus, g(x) = 2x^2+4
• Nov 5th 2012, 06:48 AM
skeeter
Re: Please help with derivatives and functions
$g''(x) = 4$

antiderivative ...

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

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

$g'(x) = 4x$

antiderivative ...

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

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

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

next time, post calculus questions in the calculus forum ... the lobby is just a place for introductions.
• Nov 5th 2012, 06:50 AM
HallsofIvy
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
• Nov 5th 2012, 07:03 AM
fkf
Re: Please help with derivatives and functions
Oh, my bad. Sorry for that.
• Nov 5th 2012, 08:40 AM
ZKR
Re: Please help with derivatives and functions
Thanks guys.
It was very helpful.

Next time I`ll post in the appropriate forum.