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?
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
a