# find f(x) dy/dx

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• Jan 11th 2009, 03:23 PM
qzno
find f(x) dy/dx
The graph of $\displaystyle y = f(x)$ passes through the point $\displaystyle (1,2)$ and has a horizontal tangent at that point.

If $\displaystyle \frac{d^2y}{dx^2} = 4x - 5$ find f(x) exactly.

I have absolutely no idea on how to even approach this question

Thank you to anyone who attempts : )
• Jan 11th 2009, 03:25 PM
Jester
Quote:

Originally Posted by qzno
The graph of $\displaystyle y = f(x)$ passes through the point $\displaystyle (1,2)$ and has a horizontal tangent at that point.

If $\displaystyle \frac{d^2y}{dx^2} = 4x - 5$ find f(x) exactly.

I have absolutely no idea on how to even approach this question

Thank you to anyone who attempts : )

Integrate once and set $\displaystyle y' = 0, x = 1$ to find your first constant of integration, then integrate again and use the fact that (1,2) is on the curve to find the second constant of integration.
• Jan 11th 2009, 03:31 PM
qzno
what do you mean by integrate and constants of integration?
• Jan 11th 2009, 03:42 PM
qzno
so i take the antiderivative of $\displaystyle 4x -5$ and i get $\displaystyle 2x^2 - 5x + c$ and then i use the info given to solve for c?
• Jan 11th 2009, 04:02 PM
vincisonfire
That's it. You have the conditions $\displaystyle y(1) =2$and$\displaystyle y'(1)=0$ Don't forget to derive twice (two constants)