Find the relative extrema, if any, of the function. Use the second derivative test, if applicable.


g(x)=4x^2+7x+6

relative maximum

(x, y) = (, )

relative minimum

(x, y) = (, )

To find the relative extrema of the function g(x) = 4x^2 + 7x + 6, we first need to find the critical points. These occur where the derivative is either zero or undefined.

Step 1: Calculate the derivative of g(x)
g'(x) = 8x + 7

Step 2: Set g'(x) equal to zero and solve for x to find the critical point(s)
8x + 7 = 0
8x = -7
x = -7/8

The only critical point for this function is x = -7/8.

Now, let's check if this critical point is a relative maximum or minimum by using the second derivative test.

Step 3: Calculate the second derivative of g(x)
g''(x) = 8

Since the second derivative is a constant (8), we can use the second derivative test to determine the nature of the critical point.

Step 4: Evaluate g''(-7/8) to determine if it is positive or negative
g''(-7/8) = 8

Since g''(-7/8) = 8 is positive, we can conclude that the critical point x = -7/8 corresponds to a relative minimum.

Therefore, the relative extrema for the function g(x) = 4x^2 + 7x + 6 are:

Relative minimum: (x, y) = (-7/8, g(-7/8))

To find the corresponding y-value, substitute x = -7/8 into the original function:

g(-7/8) = 4(-7/8)^2 + 7(-7/8) + 6
g(-7/8) = 4(49/64) - 49/8 + 6
g(-7/8) = 49/16 - 98/16 + 96/16
g(-7/8) = 47/16

Therefore, the relative minimum is: (x, y) = (-7/8, 47/16)

First you have to find the derivative of g(x). Then set that equal to 0 and solve for the value(s) of x. Then plug your critical #'s into the original equation. The largest # will be your relative max and the smallest # will be the relative min.