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Asked by Kerry

(x) = x ^3 - 12 x + 11, write the equation of the line tangent to the curve at x = 1
13 years ago

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Answered by Steve
f'(x) = 3x^2 - 12
f'(1) = -9
f(1) = 0

so you want the line through (1,0) with slope -9:

y = -9(x-1)
13 years ago
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(x) = x ^3 - 12 x + 11, write the equation of the line tangent to the curve at x = 1

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