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) = (, )
your curve is a parabola opening upwards, so there is only one minimum and no maximum196/64 + 7(
g ' (x) = 8x + 7
= 0
8x = -7
x=-7/8
g(-7/8) = 196/64 - 49/8 + 6
= 47/16
minimum or vertex is (-7/8 , 47/16)
To find the relative extrema of the function g(x) = 4x^2 + 7x + 6, we need to take the first and second derivatives of the function.
First derivative:
g'(x) = 8x + 7
Set g'(x) = 0 to find critical points:
8x + 7 = 0
8x = -7
x = -7/8
Now that we have the critical point x = -7/8, we can use the second derivative test to determine whether it is a relative maximum or minimum.
Second derivative:
g''(x) = 8
Since the second derivative, g''(x), is a constant (8), it does not change sign. Therefore, we cannot use the second derivative test to determine the nature of the critical point.
To determine if the critical point is a relative maximum or minimum, we can look at the concavity of the function. Since g''(x) is positive, the function is concave up.
Since the function is concave up and the critical point is the only point where g'(x) equals zero, the critical point x = -7/8 is a relative minimum.
Now, we can find the corresponding y-coordinate at the relative minimum by substituting x = -7/8 back into the original function:
g(-7/8) = 4(-7/8)^2 + 7(-7/8) + 6
Simplifying:
g(-7/8) = 4(49/64) - 49/8 + 6
g(-7/8) = 49/16 - 49/8 + 6
g(-7/8) = 49/16 - 98/16 + 96/16
g(-7/8) = -49/16 + 96/16
g(-7/8) = 47/16
Therefore, the relative minimum of the function g(x) = 4x^2 + 7x + 6 is at the point (x, y) = (-7/8, 47/16).