Final answer:
The function f(x)=8+6x^2+4x^3 has a local maximum at x=-1 and a local minimum at x=0. To find these, we need to solve for when the first derivative is zero to find critical points and then use the second derivative to test the nature of each critical point.
Step-by-step explanation:
In order to find the local maximum and minimum values of the function №(x)=8+6x2+4x3, we need to use both the first and second derivative tests. The first step involves calculating the derivatives.
The first derivative of the function, №'(x), will give us the critical points where the function has the potential to have a local maximum or minimum. The function's first derivative is №'(x)=12x+12x2. Setting this equal to zero, we can solve for x to find the critical points. The second derivative, №''(x), which is 12+24x, can be used to determine the concavity and identify whether we have a local maximum, minimum, or neither at these critical points.
From the first derivative, to find the critical points, solve the equation 12x+12x2=0. This simplifies to x(12+12x)=0, which has solutions x=0 and x=-1. We then use the second derivative test to determine the nature of each critical point: at x=0, №''(0)=12 which is positive, indicating a local minimum, and at x=-1, №''(-1)=12-24 which is negative, indicating a local maximum.
The local maximum value is at x=-1, and the local minimum value is at x=0 for the function №(x).