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what's the inverse of f(x)=-(x-2)^1/2+4

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3 Answers

Write y, instead of f(x), then for an inverse, exchange x and y,

So we get:    x = -(y – 2 )^1/2  + 4

                 x - 4 = -(y – 2 )^1/2   

multiply by – 1:

                4 – x = (y – 2 )^1/2   

Square both sides:

          (4 – x)^2   = y – 2

FOIL: 16 – 8x  + x^2  = y – 2

SO:             y    =    x^2  – 8x  + 16  or  y  =  (x – 4)^2
Hi Rich;
f(x)=-(x-2)^1/2+4
y=-(x-2)1/2+4
Let's subtract 4 from both sides...
y-4=-(x-2)1/2+4-4
y-4=-(x-2)1/2
Let's multiply both sides by -1...
-1(y-4)=(-1)[-(x-2)1/2]
4-y=(x-2)1/2
Let's square both sides...
(4-y)2=[(x-2)1/2]2
(4-y)2=x-2
Let's FOIL the left side...
FIRST...(4)(4)=16
OUTER...(4)(-y)=-4y
INNER...(-y)(4)=-4y
LAST...(-y)(-y)=y2
y2-8y+16=x-2
Let's add 2 to both sides...
y2-8y+16+2=x-2+2
y2-8y+18=x
Let's switch x and y...
x2-8x+18=y
x2-8x+18=f-1(x)
 
 
When doing inverses be sure to find the domain and range of the original function. They will become the range and domain of the inverse, respectively.

In this case the original function is the BOTTOM HALF of a horizontal parabola. It’s inverse will be the LEFT SIDE of a vertical parabola. (Do quick sketches using transformations of f and then reflecting over y = x. See GeoGebra sketch here:
http://www.wyzant.com/resources/files/263815/inverse_of_a_square_root_function)

Domain of f: x-2 >= 0, x >= 2
Range of f: y <= 4

Range of f^(-1): y >= 2
Domain of f^(-1): x <= 4

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

x = -(g - 2)^(1/2) + 4

x-4 = -(g - 2)^(1/2)

(x-4)^2 = g - 2

g(x) = f^(-1)(x) = (x-4)^2 + 2, x <= 4