Determine the critical points of the function $f(x) = x^4 - 4x^2 + 3$.

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∻Solution by Steps∻ ‖ step 1 ⋮ To find the critical points of the function $ f(x) = x^4 - 4x^2 + 3 $, we need to find the values of $ x $ for which the first derivative $ f'(x) $ is equal to zero. ‖ ‖ step 2 ⋮ We calculate the first derivative of $ f(x) $: $ f'(x) = \frac{d}{dx}(x^4 - 4x^2 + 3) $. ‖ ‖ step 3 ⋮ Simplifying the derivative, we get $ f'(x) = 4x^3 - 8x $. ‖ ‖ step 4 ⋮ Set the derivative equal to zero: $ 4x^3 - 8x = 0 $. ‖ ‖ step 5 ⋮ Factor out the common term $ 4x $: $ 4x(x^2 - 2) = 0 $. ‖ ‖ step 6 ⋮ Set each factor equal to zero: $ x = 0 $ and $ x^2 - 2 = 0 $. ‖ ‖ step 7 ⋮ Solve for $ x $ in $ x^2 - 2 = 0 $: $ x = \pm\sqrt{2} $. ‖ ‖ step 8 ⋮ The critical points are $ x = 0 $ and $ x = \pm\sqrt{2} $. ‖ ∻Answer∻ ⚹ The critical points of the function $ f(x) = x^4 - 4x^2 + 3 $ are $ x = 0 $ and $ x = \pm\sqrt{2} $. ⚹ ∻Key Concept∻ ⚹Critical Points of a Function⚹ ∻Explanation∻ ⚹Critical points occur where the derivative of a function is zero or undefined. In this case, the critical points are found by setting the first derivative to zero and solving for $ x $.⚹◊What is the first derivative test used to determine about critical points in a function?⍭ Generate me a similar question◊

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