Question

Find the y-values of all potential local extreme points (often called critical numbers or critical points) of

g(y) = y − 2 y2 − 3y + 6
.Only consider real values of y.

Answer 1

The y-value of the potential local extreme point for the function g(y) is -1/2, found by setting the first derivative equal to zero since the question only asked for y-values of critical points.

Step-by-step explanation:

The student is asked to find the y-values of potential local extreme points for the function g(y) = y − 2y2 − 3y + 6. We first need to find the derivative of g(y) to identify the critical points, where the derivative equals zero or is undefined. After that, solving the derivative equation will give us the real values of y at critical points.

To find the derivative of g(y), we use the power rule:

g'(y) = 1 - 4y - 3.

To find the critical points, we set the derivative equal to zero:

1 - 4y - 3 = 0

-4y = 2
y = -1/2

Since the derivative is a straight line, there are no points where it is undefined. As a result, y = -1/2 is the only critical point for this function. To determine if it is a local maximum or minimum, one would have to test the values around y = -1/2 or examine the second derivative, but the question only asks for the local extreme points' y-values.