Answer: 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
