The function provided is:

$f(x) = \frac{1}{3}x^3 - 2x$

You are asking about the function with respect to the x-coordinate $x_0 = 1$, likely looking for the function value, slope (derivative), or some related property at this point. Here's a breakdown of possible tasks:

1. Find $f(1)$:

Substitute $x = 1$ into the function:

$f(1) = \frac{1}{3}(1)^3 - 2(1)$ $f(1) = \frac{1}{3} - 2 = -\frac{5}{3}$

Thus, $f(1) = -\frac{5}{3}$.

2. Find $f'(x)$ (the derivative):

The derivative of $f(x)$ is:

$f'(x) = \frac{d}{dx}\left(\frac{1}{3}x^3 - 2x\right)$ $f'(x) = x^2 - 2$

3. Find $f'(1)$ (the slope at $x_0 = 1$):

Substitute (x