Find the local maxima of f(x) = -x^2 + 4x + 1. (2020)

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Question: Find the local maxima of f(x) = -x^2 + 4x + 1. (2020)

Options:

Correct Answer: 5

Exam Year: 2020

Solution:

The maximum occurs at x = -b/(2a) = -4/(2*-1) = 2. f(2) = -2^2 + 4(2) + 1 = 5.

Find the local maxima of f(x) = -x^2 + 4x + 1. (2020)

Find the local maxima of f(x) = -x^2 + 4x + 1. (2020)

Step 1: Identify the function we are working with, which is f(x) = -x^2 + 4x + 1.
Step 2: Recognize that this is a quadratic function in the form f(x) = ax^2 + bx + c, where a = -1, b = 4, and c = 1.
Step 3: To find the x-coordinate of the local maximum, use the formula x = -b/(2a).
Step 4: Substitute the values of a and b into the formula: x = -4/(2 * -1).
Step 5: Calculate the denominator: 2 * -1 = -2.
Step 6: Now calculate x: x = -4 / -2 = 2.
Step 7: To find the maximum value, substitute x = 2 back into the function: f(2) = -2^2 + 4(2) + 1.
Step 8: Calculate f(2): f(2) = -4 + 8 + 1 = 5.
Step 9: Therefore, the local maximum occurs at x = 2 and the maximum value is f(2) = 5.

Quadratic Functions – Understanding the properties of quadratic functions, including how to find their maxima and minima using the vertex formula.
Vertex Formula – Using the vertex formula x = -b/(2a) to find the x-coordinate of the vertex of a parabola.
Function Evaluation – Evaluating the function at the x-coordinate found to determine the maximum value.