The problem asks us to find the function $ f(x) $ given its derivative, $ f'(x) = 3x^2 - \frac{2}{x^2} $, and a specific value of the function, $ f(1) = 0 $. This requires us to perform integration and then use the given condition to find the constant of integration.
First, we find the indefinite integral of $ f'(x) $ to get the general form of $ f(x) $. We can rewrite $ f'(x) $ as $ f'(x) = 3x^2 - 2x^{-2} $.
The integral is calculated as follows:
$ f(x) = \int f'(x) dx $ $ f(x) = \int (3x^2 - 2x^{-2}) dx $Using the power rule for integration, which states that $ \int ax^n dx = a \frac{x^{n+1}}{n+1} + C $ (where $ n \neq -1 $):
$ \int 3x^2 dx = 3 \frac{x^{2+1}}{2+1} = 3 \frac{x^3}{3} = x^3 $And for the second term:
$ \int -2x^{-2} dx = -2 \frac{x^{-2+1}}{-2+1} = -2 \frac{x^{-1}}{-1} = 2x^{-1} = \frac{2}{x} $Combining these results and adding the constant of integration $ C $:
$ f(x) = x^3 + \frac{2}{x} + C $We are given that $ f(1) = 0 $. We substitute $ x = 1 $ into the expression for $ f(x) $ we found:
$ f(1) = (1)^3 + \frac{2}{(1)} + C $Since $ f(1) = 0 $, we set the expression equal to 0:
$ 0 = 1 + 2 + C $ $ 0 = 3 + C $Solving for $ C $:
$ C = -3 $Now, we substitute the value $ C = -3 $ back into the general form of $ f(x) $:
$ f(x) = x^3 + \frac{2}{x} - 3 $This is the specific function that satisfies both the derivative $ f'(x) $ and the condition $ f(1) = 0 $.