Hello,
Im 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, Id 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.
Whats the function g(x) ?

Any help is most appreciated.
Thanks

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

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

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.

Originally Posted by fkf
We have
g''(x) = 4, g'(0) = 4 and g(0) = 4
The problem gave g'(0)= 0.

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