Question
Find the points on the curve $ y = 2x^3 + 3x^2 - 12x + 1 $ where the tangent is horizontal.
Step 1
The derivative of a function gives us the slope of the tangent line at any point on the curve. So, we differentiate $ y = 2x^3 + 3x^2 - 12x + 1 $ with respect to $x$ to get: \[ y' = 6x^2 + 6x - 12 \] Show more…
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