Question

1. $f(x)=x^2 - 2x - 8$ on $-1, 3$

Explanation:

Assuming the problem is to find the extrema (maximum and minimum) of the function \( f(x) = x^2 - 2x - 8 \) on the interval \([-1, 3]\), we can solve it as follows:

Step 1: Find the critical points

First, we find the derivative of the function \( f(x) \). The derivative of \( f(x) = x^2 - 2x - 8 \) using the power rule is:
\( f'(x) = 2x - 2 \)

To find the critical points, we set \( f'(x) = 0 \) and solve for \( x \):
\( 2x - 2 = 0 \)
\( 2x = 2 \)
\( x = 1 \)

We also need to check if the derivative is undefined at any point in the interval \([-1, 3]\), but since \( f'(x) = 2x - 2 \) is a polynomial, it is defined for all real numbers, so the only critical point in the interval is \( x = 1 \).

Step 2: Evaluate the function at the critical point and the endpoints

Now we evaluate \( f(x) \) at \( x = -1 \), \( x = 1 \), and \( x = 3 \).

• For \( x = -1 \):

\( f(-1) = (-1)^2 - 2(-1) - 8 = 1 + 2 - 8 = -5 \)

• For \( x = 1 \):

\( f(1) = (1)^2 - 2(1) - 8 = 1 - 2 - 8 = -9 \)

• For \( x = 3 \):

\( f(3) = (3)^2 - 2(3) - 8 = 9 - 6 - 8 = -5 \)

Step 3: Determine the extrema

From the values we calculated:

• The minimum value of \( f(x) \) on \([-1, 3]\) is \( -9 \) (occurring at \( x = 1 \)).
• The maximum value of \( f(x) \) on \([-1, 3]\) is \( -5 \) (occurring at \( x = -1 \) and \( x = 3 \)).

Answer:

The minimum value of \( f(x) \) on \([-1, 3]\) is \(\boxed{-9}\) and the maximum value is \(\boxed{-5}\).