f(x)=1/3*x³-4x+1,x0=3

To analyze the function \( f(x) = \frac{1}{3} x^3 - 4x + 1 \) at the point \( x_0 = 3 \), we can find the value of the function and compute its derivative to understand its behavior.

Step 1: Evaluate \( f(x) \) at \( x_0 = 3 \)

\[ f(3) = \frac{1}{3} (3)^3 - 4(3) + 1 \] \[ = \frac{1}{3} (27) - 12 + 1 \] \[ = 9 - 12 + 1 \] \[ = -2 \]

Step 2: Compute the derivative \( f'(x) \)

The derivative of the function gives us the slope of the tangent line at any point \( x \):

\[ f'(x) = \frac{d}{dx} \left( \frac{1}{3} x^3 - 4x + 1 \right) \] \[ = x^2 - 4 \]

Step 3: Evaluate \( f'(x) \) at \( x_0 = 3 \)

\[ f'(3) = (3)^2 - 4 \] \[ = 9 - 4 \] \[ = 5 \]

Summary of Results

• \( f(3) = -2 \)
• \( f'(3) = 5 \)

This means at the point \( (3, -2) \), the function is decreasing, and the slope of the tangent line at that point is 5, indicating that the function is increasing at that point. If you need further analysis or information, such as finding critical points or inflection points, feel free to ask!