find an equation of the tangent line to the curve at the point given y = x^3 -3x + 1, (4,53)
Added by Michael C.
Step 1
Differentiate y = x^3 - 3x + 1 with respect to x: dy/dx = 3x^2 - 3 Evaluate the derivative at x = 4: dy/dx = 3(4)^2 - 3 dy/dx = 3(16) - 3 dy/dx = 48 - 3 dy/dx = 45 Show more…
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