Question 1199856
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<font color=red>Answer:</font> *[Tex \Large f^{-1}(\text{x}) = (\text{x}-2)^2-2]
Domain of inverse:  {{{x >= 2}}}
Range of inverse:  {{{y >= -2}}}
The graph is shown below.


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Work Shown:
*[Tex \Large f(\text{x}) = \sqrt{\text{x}+2}+2]


*[Tex \Large \text{y} = \sqrt{\text{x}+2}+2] 


*[Tex \Large \text{x} = \sqrt{\text{y}+2}+2]


*[Tex \Large \text{x}-2 = \sqrt{\text{y}+2}]


*[Tex \Large (\text{x}-2)^2 = \text{y}+2]


*[Tex \Large (\text{x}-2)^2-2 = \text{y}]


*[Tex \Large \text{y} = (\text{x}-2)^2-2]


*[Tex \Large f^{-1}(\text{x}) = (\text{x}-2)^2-2]


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Explanation:
Replace f(x) with y. Swap x and y as shown in step 3. Then solve for y to determine the inverse.


The domain of f(x) is {{{x >= -2}}} which is to avoid a negative value under the square root.
This means the range of the inverse is {{{y >= -2}}}
The domain and range swap roles when going from original function to inverse, and vice versa, simply because of the swap of x and y mentioned earlier.


The range of f(x) is {{{y >= 2}}}
This means the domain of the inverse is {{{x >= 2}}}


Graph:
{{{graph(400,400,-5,5,-5,5,-100,sqrt(x+2)+2,(sqrt(x-2)/sqrt(x-2))*((x-2)^2-2),-100,-100,-100,(sqrt(sin(5*x))/sqrt(sin(5*x)))*x))}}}
The original f(x) function is in green.
The inverse function is in blue.
To go from the green curve to the blue curve, and vice versa, reflect over the dashed line y = x.


Note that the inverse is a half-parabola since it is only drawn when {{{x >= 2}}} (domain of the inverse).


Graphing tools such as Desmos and GeoGebra are very useful as quick visual verification methods.
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