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), we need to find the x-value at which the derivative of g(x) equals zero.
1. Find the derivative of g(x):
g'(x) = 3x^2 - 12
2. Set g'(x) equal to zero and solve for x:
3x^2 - 12 = 0
Divide both sides of the equation by 3:
x^2 - 4 = 0
3. Factor the quadratic equation:
(x - 2)(x + 2) = 0
The solutions are x = 2 and x = -2.
Now, we need to determine which solution corresponds to the local minimum. To do this, we can use the second derivative test.
4. Find the second derivative of g(x):
g''(x) = 6x
5. Substitute the x-values into g''(x):
g''(2) = 6(2) = 12
g''(-2) = 6(-2) = -12
Since the second derivative at x = 2 is positive (g''(2) > 0), this means that x = 2 corresponds to a local minimum.
Now, let's evaluate g(x) at the x-value of the ordered pair location of this local minimum.
6. Substitute x = 2 into g(x):
g(2) = (2)^3 - 12(2) + 16
g(2) = 8 - 24 + 16
g(2) = 0
Therefore, the location of the local minimum is (2, 0), and the value of g(x) at this location is 0.
To find the location of the local minimum of the function g(x) = x^3 - 12x + 16, we need to find the critical points first.
Step 1: Calculate the derivative of g(x) with respect to x:
g'(x) = 3x^2 - 12
Step 2: Set g'(x) equal to zero and solve for x to find the critical points:
3x^2 - 12 = 0
Step 3: Factor out a common factor from the equation:
3(x^2 - 4) = 0
Step 4: Set each factor equal to zero:
x^2 - 4 = 0
Step 5: Solve for x:
x^2 = 4
Taking the square root of both sides: x = ±2
So, the critical points are x = -2 and x = 2.
To determine if these critical points are local minima or maxima, we need to evaluate the second derivative of g(x).
Step 6: Calculate the second derivative of g(x) with respect to x:
g''(x) = (3x^2 - 12)' = 6x
Step 7: Evaluate g''(x) at the critical points:
g''(-2) = 6(-2) = -12
g''(2) = 6(2) = 12
If g''(x) is positive at a critical point, it indicates a local minimum, and if it is negative, it indicates a local maximum.
Since g''(-2) = -12 < 0, the critical point x = -2 corresponds to a local maximum.
On the other hand, g''(2) = 12 > 0, which means the critical point x = 2 corresponds to a local minimum.
Now we can find the y-value of the local minimum by substituting x = 2 into the original function g(x).
g(2) = (2)^3 - 12(2) + 16
= 8 - 24 + 16
= 0
Therefore, the x-coordinate of the local minimum is 2, and the corresponding y-coordinate is 0.