Find the values of x that give relative extrema for the function f(x) = (x + 1 )^2 (x -2).?

Question

Can you please Show me Steps for this. I need to Know how to do it.

Answer 1

The relative extrema for the functions are found by
taking the first derivative and setting it equal to zero
f'(x)= 2(x+1)(1)(x-2)+(x+1)^2(1), using the chain and the product rule
f'(x) = 2(x+1)(x-2)+(x+1)^2
0 = 2(x+1)(x-2)+(x+1)^2
solving for x, one obtains
x = +- 1 for answer

the x values are +- 1 and these are the values in which the graph of f attains either a maximum or a minimum

Answer 2

f(x) = (x + 1)²(x - 2)
f(x) = (x² + 2x + 1)(x - 2)
f(x) = x^3 - 3x - 2

extrema when f'(x) = 0

f'(x) = 3x² - 3 = 0
3x² = 3
x² = 1
x = ±1