Question

if f(x) = sin x 2x 1 and g is the inverse function of f, what is the value of g'(1) ?

Answer 1

Using inverse functions, it is found that:

$g^{-1}(1) = k\frac{\pi}{2}, k = \pm 1, \pm 3, \pm 5, ...$

The function given is:

$f(x) = y = \sin{x} + 1$

To find the inverse, we exchange x and y, and isolate y, hence:

$x = \sin{y} + 1$

$x - 1 = \sin{y}$

$y = \arcsin{(x - 1)}$

Then:

$g^{-1}(x) = \arcsin{(x - 1)}$

At x = 1:

$g^{-1}(1) = \arcsin{(1 - 1)} = k\frac{\pi}{2}, k = \pm 1, \pm 3, \pm 5, ...$

To learn more about inverse functions, you can take a look at https://brainly.com/question/25897382