`f(x) = x^3 + 1, (-1, 0)` Find an equation of the tangent line to the graph of f at the given point.

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Chapter 2, 2.1 - Problem 28 - Calculus of a Single Variable (10th Edition, Ron Larson).
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hkj1385 | (Level 1) Assistant Educator

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The given function is:-

f(x) = (x^3) + 1

differentiating both sides w.r.t 'x' we get

f'(x) = 3*(x^2) + 0

Now, slope of the tangent at the point (-1,0) = f'(-1) = 3*(-1^2) = 3

Thus, equation of the tangent at the point (-1,0) and having slope = 3 is :-

y - 0 = (3)*(x - (-1))

or, y = 3x + 3

or, y - 3x = 3 is the equation of the tangent to the  given curve at (-1,0)

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