To determine the local minimum or maximum of a function, we first need to find the critical points of the function which are points where the derivative of the function equals to 0 or is undefined.
First, let's find the derivative of the function f(x) = 2x³-24x² +72x+8.
The derivative, f'(x), of the function is 6x² - 48x + 72.
Setting f'(x) equal to 0, we solve for x:
6x² - 48x + 72 = 0
Divide each term by 6 to simplify:
x² - 8x + 12 = 0
This equation factors to
(x-2)(x-6) = 0
Setting each factor equal to zero gives the solutions x=2 and x=6 which are the critical points.
Next we will apply the second derivative test to classify the critical points. First we find the second derivative:
f''(x) = 12x - 48
Let's evaluate the second derivative at each critical point:
For x=2: f''(2) = 24 - 48 = -24 which is less than 0, thus at x=2 we have a local maximum.
For x=6: f''(6) = 72 - 48 = 24 which is greater than 0, thus at x=6 we have a local minimum.
Finally, we substitute x=2 and x=6 into the original function to find the y-values for these points:
f(2) = 2(2)³ - 24(2)² + 72(2) + 8 = 16 - 96 + 144 + 8 = 72
f(6) = 2(6)³ - 24(6)² + 72(6) + 8 = 432 - 864 + 432 + 8 = 8
So, the function has a local maximum at x = 2 with a value 72, and a local minimum at x = 6 with a value 8.