Question

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 ) =

Answer 1

Answer:

  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

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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.

Answer 2

Using function concepts, it is found that:

• The domain on which f is one-to-one and non-decreasing is: $x \geq -3$.
• The inverse function restricted to this domain is $f^{-1}(x) = \sqrt{x} - 3$.

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Suppose we have a parent square function:

$f(x) = g(x)^2$

• It is one-to-one for either: $g(x) \geq 0$ or $g(x) < 0$
• It is non-decreasing for: $g(x) \geq 0$

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In this question, we have:

$f(x) = (x + 3)^2$

• It has $g(x) = x + 3$
• It is one-to-one and non-decreasing for: $g(x) \geq 0$, thus:

$x + 3 \geq 0$

$x \geq -3$.

The domain on which f is one-to-one and non-decreasing is: $x \geq -3$.

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

$y = (x + 3)^2$

$x = (y + 3)^2$

$\sqrt{(y+3)^2} = \sqrt{x}$

$y + 3 = \sqrt{x}$

$y = \sqrt{x} - 3$

$f^{-1}(x) = \sqrt{x} - 3$

The inverse function restricted to this domain is $f^{-1}(x) = \sqrt{x} - 3$

A similar problem is given at https://brainly.com/question/23826461