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$y = x^3 - 6x^2 + x + 3$. The tangent is parallel to the line $y = x + 5$ at points where $x$ equals:
A0 and 4
B0 and 6
CBoth roots of $3x^2 - 12x = 0$, i.e. $x = 0, 4$
D$x = 2$ only
Answer & Solution
Correct answer: C. Both roots of $3x^2 - 12x = 0$, i.e. $x = 0, 4$
Tangent parallel to $y = x + 5$ has slope 1. $y' = 3x^2 - 12x + 1 = 1$ ⇒ $3x^2 - 12x = 0$ ⇒ $x = 0$ or $x = 4$.
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