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 first need to find its derivative. ‖ ‖ step 2 ⋮ The derivative of $f(x)$ is $f'(x) = 3x^2 - 3$. ‖ ‖ step 3 ⋮ Set the derivative equal to zero to find the critical points: $3x^2 - 3 = 0$. ‖ ‖ step 4 ⋮ Solve the equation $3x^2 - 3 = 0$ for $x$. ‖ ‖ step 5 ⋮ Simplify the equation: $3(x^2 - 1) = 0$. ‖ ‖ step 6 ⋮ Factor the equation: $3(x - 1)(x + 1) = 0$. ‖ ‖ step 7 ⋮ Set each factor equal to zero: $x - 1 = 0$ and $x + 1 = 0$. ‖ ‖ step 8 ⋮ Solve for $x$: $x = 1$ 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⚹ ∻Explanation∻ ⚹Critical points of a function occur where its derivative is zero or undefined. In this case, we found the points where the derivative $f'(x) = 3x^2 - 3$ is zero.⚹◊What is the formula for finding the critical points of a function in calculus?⍭ Generate me a similar question◊

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