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Let g: R -> R be defined by g(x) = e^(2x) + 3x + sin(x) + 1. If g^(-1) denotes the inverse function of g, find the value of 1 / (g^(-1))'(2).

1. 6
2. 1/6
3. 5
4. 1/5

Correct answer: 6

Solution

First find x0 = g^(-1)(2): g(x0) = 2. Try x0=0: g(0) = e⁰ + 0 + sin(0) + 1 = 1+0+0+1 = 2. Yes, x0=0. By the inverse function theorem: (g^(-1))'(2) = 1/g'(0). g'(x) = 2e^(2x) + 3 + cos(x). g'(0) = 2*1 + 3 + 1 = 6. Therefore 1/(g^(-1))'(2) = g'(0) = 6.

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