Find the slope of the tangent to the curve $y = 3x^2 - 2x + 1$ at $x = 1$.

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∻Solution by Steps∻ ‖ step 1 ⋮ To find the slope of the tangent line to the curve $y = 3x^2 - 2x + 1$ at $x = 1$, we first need to find the derivative of the function. ‖ ‖ step 2 ⋮ The derivative of $y = 3x^2 - 2x + 1$ is $y' = \frac{d}{dx}(3x^2 - 2x + 1)$. ‖ ‖ step 3 ⋮ Using the power rule, we get $y' = 6x - 2$. ‖ ‖ step 4 ⋮ Now, we evaluate the derivative at $x = 1$: $y'(1) = 6(1) - 2 = 4$. ‖ ‖ step 5 ⋮ Therefore, the slope of the tangent line to the curve at $x = 1$ is $4$. ‖ ∻Answer∻ ⚹ The slope of the tangent line to the curve $y = 3x^2 - 2x + 1$ at $x = 1$ is $4$. ⚹ ∻Key Concept∻ ⚹Derivative⚹ ∻Explanation∻ ⚹The derivative of a function at a given point gives the slope of the tangent line to the curve at that point.⚹◊What is the general formula for finding the slope of a tangent line to a curve at a given point⍭ Generate me a similar question◊

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