The problem is "If f(x)=sinx+2x+1 and g is the inverse of f, what is the value of g'(1) (the derivative of g at x=1)?"

I have no clue how to inverse this. What is the solution?

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The problem is "If f(x)=sinx+2x+1 and g is the inverse of f, what is the value of g'(1) (the derivative of g at x=1)?"

I have no clue how to inverse this. What is the solution?

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## Answers

y = sin(x) + 2x + 1

x = sin(y) + 2y + 1

x = sin(g(x)) + 2 * g(x) + 1

Derive implicitly and solve for g'(x)

1 = g'(x) * cos(g(x)) + 2 * g'(x)

1 = g'(x) * (2 + cos(g(x))

g'(x) = 1 / (2 + cos(g(x))

g'(1) = 1 / (2 + cos(g(1))

x = sin(g(x)) + 2 * g(x) + 1

1 = sin(g(1)) + 2 * g(1) + 1

0 = sin(g(1)) + 2 * g(1)

-2 * g(1) = sin(g(1))

This only happens when g(1) = 0

g'(1) = 1 / (2 + cos(g(1))

g'(1) = 1 / (2 + cos(0))

g'(1) = 1 / (2 + 1)

g'(1) = 1/3

EDIT:

Just for fun, let's see how this corresponds to f'(0)?

f(x) = sin(x) + 2x + 1

f'(x) = cos(x) + 2

f'(0) = 1 + 2

f'(0) = 3

So, it appears that for inverse functions f(x) and g(x), f'(x) * g'(x) = 1. It'd be worth investigating in order to determine if that holds true all the time.

f(x) = sin x + 2x + 1

f^-1(x) = g(x)

Let f^-1(1) = c

f(c) = 1

g(1) = c

sin c + 2c + 1 = 1

sin c + 2c = 0

c= 0 is a solution

g(1) = 0

g'(1) = 1/ f'( g(1) )

g'(1) = 1 / f'(0)

f'(x) = cos x + 2

f'(0) = cos (0) + 2

f'(0) = 3

g'(1) = 1/3

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