Let g(x) = x3 - 12x + 16. Find the location of the local minimum of this function (there is only one such location). Evaluate function g(x) at the x-value of the ordered pair location of this local minimum.
To find the location of the local minimum of the function g(x) = x^3 - 12x + 16, we can start by taking the derivative of the function and setting it equal to zero. This will allow us to find the critical points of the function, where the slope is either zero or undefined.
1. Take the derivative of g(x):
g'(x) = 3x^2 - 12
2. Set the derivative equal to zero and solve for x to find the critical points:
3x^2 - 12 = 0
Add 12 to both sides:
3x^2 = 12
Divide both sides by 3:
x^2 = 4
Take the square root of both sides:
x = ±2
We have found two critical points, x = -2 and x = 2.
3. To determine which critical point corresponds to a local minimum, we can analyze the second derivative of the function g(x).
Take the derivative of g'(x):
g''(x) = 6x
Evaluate g''(x) at the critical points:
g''(-2) = 6(-2) = -12
g''(2) = 6(2) = 12
Since g''(-2) is negative and g''(2) is positive, we can conclude that x = 2 corresponds to a local minimum.
4. Evaluate the function g(x) at x = 2 to find the value of the function at the local minimum:
g(2) = (2^3) - 12(2) + 16
= 8 - 24 + 16
= 0
Therefore, the location of the local minimum is (2, 0) and the function g(x) evaluated at this x-value is 0.