what is the inverse of 2 square root sign x-1 +3......Please explain how you do this?

Answer 1

if that equals 0 then all u gotta do is move 3 to other side by subtracting 3 and then divide both sides by 2 to move the 2 over. Then, square both sides. You get x-1 = -3/2 now add 1 to both sides and x = -1/2 plug it back in to original equation and see if it works (sometimes it doesn't when the problem involves square roots/squares).

Answer 2

f(x) = 2sqrt(x - 1) + 3

Before we solve for the inverse of this, it's best that we solve for the range of this function. After all, the range of f(x) is going to be the domain of f^(-1)(x).

Since sqrt(x - 1) >= 0, then 2sqrt(x - 1) >= 0, so adding 3 to both sides of the equation,
2sqrt(x - 1) + 3 >= 3

Therefore, the range of this function in interval notation is
y E [3, infinity)

To solve for the inverse of this, first make f(x) equal to y.

y = 2sqrt(x - 1) + 3

Now, swap the x and y variables and then solve for y.

x = 2sqrt(y - 1) + 3

Move the 3 to the left hand side,

x - 3 = 2sqrt(y - 1)

Square both sides,

(x - 3)^2 = 4(y - 1)

Expand the right hand side,

(x - 3)^2 = 4y - 4

Move the -4 to the left hand side, and then divide both sides by 4.

y = [(x - 3)^2 + 4] / 4

At this point, you need your concluding statement; change y into f^(-1)(x).

f^(-1)(x) = [(x - 3)^2 + 4] / 4

This is a parabola opening upward, and parabolas are not one-to-one functions, meaning this shouldn't have an inverse. That's what finding the range of f(x) was for; the range of f(x) will be this function's domain. Therefore, the PROPER inverse will be

f^(-1)(x) = [(x - 3)^2 + 4] / 4, for x E [3, infinity).

** The fact that the domain of this function is 3 to infinity is EXTREMELY important to include. Remember not to forget it. **

Answer 3

To find the inverse, we solve the below equation for y:
x = 2 √(y-1) + 3
=> 2 √(y-1) = x - 3
=> √(y-1) = (x-3)/2
=> y - 1 = (x-3)^2 / 4
=> y = (x-3)^2 / 4 + 1.

Answer 4

y = 2 sqrt (x-1) + 3

To find the inverse, you exchange the x and y and then solve for y.

x = 2 sqrt (y-1) + 3
x + 3 = 2 sqrt (y-1)
(x+3)/2 = sqrt (y-1)
[(x+3)/2]^2 = y - 1
[(x+3)^2] / 4 = y - 1
1 + [(x+3)^2] / 4 = y