Let f ( x ) = ( x + 3 ) ^2
Find a domain on which f is one-to-one and non-decreasing.
=
Find the inverse of f restricted to this domain
f − 1 ( x ) =

Using function concepts, it is found that:

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Suppose we have a parent square function:

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In this question, we have:

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The domain on which f is one-to-one and non-decreasing is: .

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To find the inverse function on this domain, we exchange x and y, and isolate y. Thus:

The inverse function restricted to this domain is

A similar problem is given at link

  f(x) restricted to x in [-3, ∞) will have inverse function f⁻¹(x) = √x -3

Step-by-step explanation:

The given function is non-decreasing on the interval [-3, ∞), so that is a suitable domain.

The inverse function can be found from ...

  x = f(y) = (y +3)^2

  √x = y +3

  √x -3 = y

So, the inverse function of f(x) on the domain [-3, ∞) is ...

  f⁻¹(x) = √x -3

In the attached graph, the solid red curve is the domain-restricted portion of f(x). The blue curve is the inverse function. The dashed orange line is a reference line to help you see that the inverse function is a reflection of the function across that line.

x= -3

step-by-step explanation:

step 1: simplify brackets

4 - x - 1 = 6

step 2: simpilfy

3 - x = 6

step 3: subtract 3 from both sides

-x = 6 - 3

step 4: simpilfy 6 - 3

-x = 3

step 5: multiply both sides by -1

x = -3

-22

times both sides by 2, times () by 8, add alile terms, move terms, add alike and caculate divide both sides by 3