How do you solve this problem?

Question

Find the x-values where f ' (x) = 0.

f(x)= (1/3)x^3 - x + 1

Answer 1

f(x) = (1/3)x³ - x + 1

Diferenciate each side with respective x  for function : f(x) = (1/3)x³ - x + 1

Recall: derivative of  x^n  = d/dx (xⁿ) = nx^n-1  Derivative of constant  = 0

There fore f¹(x) = (1/3)3x^3-1 - 1x^1-1 + 0

Simplify

f¹(x) = x² - x⁰ + 0

Recall: x⁰ = 1

f1(x) = x² - 1

But f¹(x) = 0

Substitute f¹(x) = 0

x² - 1 = 0

Add 1 to each side

x² -1 +1 = 0 +1

x² + 0 = 1

x² = 1

Take square root each side

x = -1, 1

x values is -1, 1.