Question

in order to find the coordinates of the local maximum point, we must find a critical point do this?
set the 2nd derivative equal to zero and solve for x.
set the function equal to zero and solve for x.
set the derivative equal to zero and solve for x.
take the derivative of the function.
answer: y=
then set the derivative equal to zero and solve for x.
answer: x =
what are the coordinates of the local maximum point? give your answer as a point: (x - value, y - value
answer:

Explanation:

Step1: Identify the function (missing in the image, assume a typical function for demonstration, e.g., \( y = -x^2 + 4x - 3 \))

To find the derivative, use power rule. For \( y = ax^n \), \( y' = nax^{n - 1} \). So for \( y = -x^2 + 4x - 3 \), \( y' = -2x + 4 \).

Step2: Set derivative to zero

Set \( y' = 0 \), so \( -2x + 4 = 0 \). Solve for \( x \): \( -2x = -4 \), \( x = 2 \).

Step3: Find y - value

Substitute \( x = 2 \) into original function: \( y = -(2)^2 + 4(2) - 3 = -4 + 8 - 3 = 1 \). So the local maximum point is \( (2, 1) \). (Note: This is a demonstration; actual function from image is needed for precise answer. But the process is: take derivative, set to zero, solve for x, substitute x into original function for y.)

Answer:

(Assuming the function leads to \( x = 2 \), \( y = 1 \)) \((2, 1)\) (Replace with actual values from the function in the image)