Question

find the x-values of all points where the function has any relative extrema. find the value(s) of any relative extrema.

f(x) = -x² - 5x + 1

select the correct choice below and, if necessary, fill in any answer boxes within your choice.

a. there are no relative maxima. the function has a relative minimum of at x = .
(use a comma to separate answers as needed.)

b. there are no relative minima. the function has a relative maximum of at x = .
(use a comma to separate answers as needed.)

c. the function has a relative maximum of at x = and a relative minimum of at x = .
(use a comma to separate answers as needed.)

d. there are no relative extrema.

Explanation:

Step1: Identify the function type

The function \( f(x) = -x^2 - 5x + 1 \) is a quadratic function. The general form of a quadratic function is \( f(x) = ax^2 + bx + c \), where \( a = -1 \), \( b = -5 \), and \( c = 1 \). Since \( a = -1 < 0 \), the parabola opens downward, so the function has a relative maximum (and no relative minimum).

Step2: Find the x-coordinate of the vertex

For a quadratic function \( f(x) = ax^2 + bx + c \), the x-coordinate of the vertex (which is the location of the relative extremum) is given by \( x = -\frac{b}{2a} \). Substituting \( a = -1 \) and \( b = -5 \) into the formula:

$$ x = -\frac{-5}{2(-1)} = -\frac{5}{2} = -2.5 $$

Step3: Find the value of the relative maximum

Substitute \( x = -2.5 \) into the function \( f(x) = -x^2 - 5x + 1 \):

$$ f(-2.5) = -(-2.5)^2 - 5(-2.5) + 1 = -6.25 + 12.5 + 1 = 7.25 = \frac{29}{4} $$

Answer:

B. There are no relative minima. The function has a relative maximum of \(\frac{29}{4}\) (or \(7.25\)) at \(x = -\frac{5}{2}\) (or \(x = -2.5\)).