Identify the function's local and absolute extreme values, if any, saying where they occur. f(x) = x^3 + 4x^2 - 3x - 4 A) local maximum at x = -1/3; local minimum at x = 3 B) local maximum at x = -1; local minimum at x = 1 C) local maximum at x = -3; local minimum at x 1/3 D) local maximum at x = -1; local minimum at x = 1

C) local maximum at x = -3; local minimum at x = 1/3

Step-by-step explanation:

Given the function:

f(x) = x^3 + 4x^2 - 3x - 4

Set f'(x) = 0

f'(x) = 3x² + 8x - 3

f'(x) = 0

3x² + 8x - 3 = 0

Using the quadratic function calculator :

x = - 3 or x = 1/3

Critical points = - 3 ; 1/3

To find the maximum and minimum points :

f''(x) = 6x + 8

Substitute the values of X in

At x = - 3

6(-3) + 8

-18 + 8 = - 10

At x = 1/3

6(1/3) + 8

2 + 8 = 10

Value of f''(x) at - 3 is negative, hence - 3 is the maximum point

Value of f''(x) at 1/3 is positive , hence 1/3 is the minimum point

the answer is the third / last answer

step-by-step explanation: