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 $f(x)$ with respect to $x$: $f'(x) = 3x^2 - 3$. ‖ ‖ 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 ⋮ Apply the zero product property: $x^2 - 1 = 0$. ‖ ‖ 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 and solve for $x$: $x + 1 = 0$ or $x - 1 = 0$. ‖ ‖ step 8 ⋮ The solutions are $x = -1$ and $x = 1$. These are the critical points of the function. ‖ ∻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 occur where the derivative of the function is zero or undefined. In this case, setting the first derivative equal to zero and solving for $x$ gives the critical points.⚹◊What is the formula for finding the critical points of a function?⍭ Generate me a similar question◊

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