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!