f(x)=10x3 + x2 - 10x, xo = -1

To analyze the function \( f(x) = 10x^3 + x^2 - 10x \) at the point \( x_0 = -1 \), we can perform a few tasks, such as calculating the value of the function at \( x_0 \), finding the derivative, and evaluating whether the point is a critical point or examining other characteristics.

Step 1: Calculate \( f(-1) \)

Let's substitute \( x = -1 \) into the function:

\[ f(-1) = 10(-1)^3 + (-1)^2 - 10(-1) \]

Calculating each term:

• \( 10(-1)^3 = 10 \cdot (-1) = -10 \)
• \( (-1)^2 = 1 \)
• \( -10(-1) = 10 \)

Now, combining the results:

\[ f(-1) = -10 + 1 + 10 = 1 \]

So, \( f(-1) = 1 \).

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

Next, we should find the derivative of \( f(x) \):

\[ f'(x) = \frac{d}{dx} (10x^3) + \frac{d}{dx} (x^2) - \frac{d}{dx} (10x) \] \[ f'(x) = 30x^2 + 2x - 10 \]

Step 3: Evaluate at \( x_0 = -1 \)

Now, we can evaluate the derivative at \( x_0 = -1 \):

\[ f'(-1) = 30(-1)^2 + 2(-1) - 10 \] Calculating each term:

• \( 30(-1)^2 = 30 \cdot 1 = 30 \)
• \( 2(-1) = -2 \)
• \( -10 \) is as it is

Now, combining the results:

\[ f'(-1) = 30 - 2 - 10 = 18 \]

Summary of Findings

• The value of the function at \( x_0 = -1 \) is \( f(-1) = 1 \).
• The derivative at \( x_0 = -1 \) is \( f'(-1) = 18 \).

Since \( f'(-1) \neq 0 \), \( x_0 = -1 \) is not a critical point. The positive value suggests that the function is increasing at this point.

If you have any specific questions or need further analysis, feel free to ask!