yes, but its not that simple UNLESS the graph is a simple polynomial.
If f'(x) is a constant i.e f'(x) = K ( a horizontal line) then f(x) = Kx + C where C is an unknown constant. Only if you are given more information can you figure out what C is.
If f'(x) = mx + b ( a non horizontal line) Un - differentiate that line and get f(x) = (m/2)x exp 2 + bx + C
Where C is an unknown constant
If f'(x) is a parabola you are going to have to figure out the equation of the parabola, before you can work backwards. All quadratics are parabolas. You should get a quadratic equation and when you work backwads you should have a cubic equation
If f'(x) is a constant i.e f'(x) = K ( a horizontal line) then f(x) = Kx + C where C is an unknown constant. Only if you are given more information can you figure out what C is.
If f'(x) = mx + b ( a non horizontal line) Un - differentiate that line and get f(x) = (m/2)x exp 2 + bx + C
Where C is an unknown constant
If f'(x) is a parabola you are going to have to figure out the equation of the parabola, before you can work backwards. All quadratics are parabolas. You should get a quadratic equation and when you work backwads you should have a cubic equation
-
I think