What is f(x) when $$f(1)=0$$
first derivative $$f(1)=1/2$$
second derivative $$f(x)=1/x^3$$
Currently i have tried where the second derivitive = first derivitive + constant at x=1
1/x^3 = f^i(x) + C 1/1 = 1/2 + C C=1/2
As well as where f(x) = first derivitve + constant at x=1
0 = x/2 + C C = -x/2
I dont know where to go from here, aswell dont know if i had started it correctly.
$f'(x) = \displaystyle \int \dfrac{1}{x^3} dx = -\dfrac{1}{x^2} + C$.
$f'(1) = \dfrac{1}{2} = -1 + C \Rightarrow C = \dfrac{3}{2}$.
$f(x) = \displaystyle \int \left(-\dfrac{1}{x^2} + \dfrac{3}{2}\right) dx = \dfrac{1}{x} + \dfrac{3}{2}x + D$.
$f(1) = 0 \Rightarrow 0 = 1 + \dfrac{3}{2} + D \Rightarrow D = -\dfrac{5}{2} \Rightarrow f(x) = \dfrac{1}{x} + \dfrac{3x}{2} - \dfrac{5}{2}$.