Question

f(x) = x^3 - 5x - 6, and f(3) = 6. Let g(x) be the inverse function of f(x). g'(6) = \frac{1}{22} Write the equation of the tangent line to g(x) at x = 6. y = 0.0454545455(x - 6) + 3

          f(x) = x^3 - 5x - 6, and f(3) = 6.
Let g(x) be the inverse function of f(x).
g'(6) = \frac{1}{22}
Write the equation of the tangent line to g(x) at x = 6.
y = 0.0454545455(x - 6) + 3

f(x) = x^3 - 5x - 6, and f(3) = 6.
Let g(x) be the inverse function of f(x).
g'(6) = (1)/(22)
Write the equation of the tangent line to g(x) at x = 6.
y = 0.0454545455(x - 6) + 3

Added by Vicente W.

Calculus: Early Transcendentals

James Stewart 8th Edition

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We are given that f(3) = 6. This means that when x = 3, the value of f(x) is 6. So we can substitute x = 3 into the equation f(x) = x^3 - 5x - 6 and solve for the constant term: f(3) = 3^3 - 5(3) - 6 = 27 - 15 - 6 = 6 Show more…

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fx = x^3 - 5x - 6, and f(3) = 6 Let g be the inverse function of f. 1. g(6) = 22 Write the equation of the tangent line to g at x = 6: y = 0.0454545455x - 6 + 3

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f(x) = x^3 - 5x - 6, and f(3) = 6. Let g(x) be the inverse function of f(x). g'(6) = \frac{1}{22} Write the equation of the tangent line to g(x) at x = 6. y = 0.0454545455(x - 6) + 3

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