if one is given the first derivative f'(x) and the graph of f passes through the point (x,y) how does one go about finding the parent function :f(x)

Question

if one is given the first derivative f'(x)  and the graph of f passes through the point (x,y) how does one go about finding the parent function :f(x)

whpalmer4 · Answer

Integrate the first derivative, then use the known point to find the constant of integration.

For example, if f'(x) = x, and a known point on f(x) is (0,2), we integrate f'(x) dx to get x^2/2 + C, then plug x = 0 into x^2/2 + C and find C such that the value of the expression is 2.  y = x^2/2 + 2 would be the parent function f(x).

anonymous · Answer

so i'm givin f'(x)=6x^2 and the graph of g passes through the point P(3,7) what would I plug in to get f(x)

whpalmer4 · Answer

If f'(x) = 6x^2, what does the indefinite integral of 6x^2 dx equal?  Isn't that going to be something involving x^3? Plug x = 3 into that formula and adjust the constant so that the result is 7.

anonymous · Answer

$$f'(x) = 6x^{2} = \int\limits_{}^{}6x^{2} dx = 2x^{3} + c $$
$$7=2(3)^{3}+c $$
Solve for c which gives you -47=c and add this to the integrated equation.

anonymous · Answer

$$f(x)=f'(x)$$

$$F'(x)=f(x)$$

$$\int f(x) dx=F(x)+c$$

^ this is the family of functions. Use your point to find what c is equal to

anonymous · Answer

to get the parent

anonymous · Answer

so f(x)= 2x^3 -47 ... right?

anonymous · Answer

sounds right

whpalmer4 · Answer

As a check:

$$f(3) = 2(3)^3 - 47 = 7  \checkmark$$$$f'(x) = 3*2x^2 - 0 = 6x^2 \checkmark$$
Looks like you got it right.