Help me please!!!I need your help immediately!!!!?

Question

Here is the function f(x)=2sinx+1 xε(0,π/2)

Find the f^-1(2) i mean the f inverse!!!!!!!!!!!

Answer 1

f(x)=2sinx+1 xε(0,π/2)
Let y=2sinx+1

2sinx=y-1
sinx=(y-1)/2
x=asin(y-1)/2 1<=y<=3
but xε(0,π/2)

so
x=asin(y-1)/2 1<=y<=3 & asin(y-1) <= π/2
and
=π/2 - asin(y-1)/2 1<=y<=3 & asin(y-1) > π/2

Therefore

f-1(x) =
x=asin(y-1)/2 1<=y<=3 & asin(y-1) <= π/2
and
=π/2 - asin(y-1)/2 1<=y<=3 & asin(y-1) > π/2

Answer 2

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

One way we can solve this is to just find the inverse. Make f(x) into y,

y = 2sin(x) + 1

Swap the x and y variables, and then solve for y.

x = 2sin(y) + 1

x - 1 = 2sin(y)
(1/2)(x - 1) = sin(y). Therefore,

y = arcsin( (1/2)(x - 1) )

Therefore,

f^(-1)(x) = arcsin ( (1/2)(x - 1) )

We want to find f^(-1)(2), so

f^(-1)(2) = arcsin ( (1/2)(2 - 1) )
= arcsin ( (1/2)(1) )
= arcsin (1/2)

Which is essentially asking where sin(x) = 1/2 on the interval (0 to &pi/2), also known as the first quadrant. On the interval from 0 to 2&pi, this would be true at the points &pi/6 and 5&pi/6. Since we're restricted only to the first quadrant, this is true at the point &pi/6, so

f^(-1)(2) = &pi/6

Answer 3

f^-1(2) mean we need to find x when f(x) = 2

So
2sinx+1 = 2
sin x = 1/2
x = Pi/6

Answer 4

Look at your previous question. I've answered it there with working.
f^-1 (2) = pi/6

Answer 5

f^-1(x) = arcsin((x - 1)/2)
f^-1(2) = arcsin((2 - 1)/2)
f^-1(2) = π/6, 5π/6

Answer 6

pi/6