Find the critical points of the function $f(x) = x^3 - 3x + 1$.

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∻Solution by Steps∻ ‖ step 1 ⋮ To find the critical points of the function $ f(x) = x^3 - 3x + 1 $, we need to find the values of $ x $ where the first derivative $ f'(x) $ is equal to zero or undefined. ‖ ‖ step 2 ⋮ First, we calculate the derivative of $ f(x) $. The derivative of $ x^3 $ is $ 3x^2 $, the derivative of $ -3x $ is $ -3 $, and the derivative of a constant is 0. So, $ f'(x) = 3x^2 - 3 $. ‖ ‖ step 3 ⋮ Set the derivative equal to zero to find critical points: $ 3x^2 - 3 = 0 $. ‖ ‖ step 4 ⋮ Factor out the common factor of 3: $ 3(x^2 - 1) = 0 $. ‖ ‖ step 5 ⋮ Set the factored expression equal to zero: $ x^2 - 1 = 0 $. ‖ ‖ step 6 ⋮ Solve the equation $ x^2 - 1 = 0 $ by factoring it as $ (x - 1)(x + 1) = 0 $. ‖ ‖ step 7 ⋮ Set each factor equal to zero: $ x - 1 = 0 $ and $ x + 1 = 0 $. ‖ ‖ step 8 ⋮ Solve for $ x $ in each equation: $ x = 1 $ and $ x = -1 $. ‖ ∻Answer∻ ⚹ The critical points of the function $ f(x) = x^3 - 3x + 1 $ are $ x = 1 $ and $ x = -1 $. ⚹ ∻Key Concept∻ ⚹Critical Points of a Function⚹ ∻Explanation∻ ⚹Critical points of a function occur where its derivative is zero or undefined. In this case, the critical points are found by setting the first derivative of the function equal to zero and solving for $ x $.⚹◊What is the formula for finding the critical points of a function in calculus?⍭ Generate me a similar question◊

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