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31 December, 10:12

Consider the function and its inverse. and When comparing the functions using the equations, which conclusion can be made? The domain of f (x) is restricted to x ≥ 0, and the domain of f-1 (x) is restricted to x ≥ 0. The domain of f (x) is restricted to x ≥ 0, and the domain of f-1 (x) is restricted to x ≤ 0. The domain of f (x) is restricted to x ≤ 0, and the domain of f-1 (x) is restricted to x ≥ 4. The domain of f (x) is restricted to x ≤ 0, and the domain of f-1 (x) is restricted to x ≤ 4.

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  1. 31 December, 10:26
    0
    Step-by-step explanation:

    The original function / displaystyle f/left (x/right) = {/left (x - 4/right) }^{2}f (x) = (x-4)

    2

    is not one-to-one, but the function is restricted to a domain of / displaystyle x/ge 4x≥4 or / displaystyle x/le 4x≤4 on which it is one-to-one.

    Two graphs of f (x) = (x-4) ^2 where the first is when x>=4 and the second is when x<=4.

    Figure 5

    To find the inverse, start by replacing / displaystyle f/left (x/right) f (x) with the simple variable y.























    y

    =

    (

    x

    -

    4

    )

    2

    Interchange

    x

    and

    y

    .

    x

    =

    (

    y

    -

    4

    )

    2

    Take the square root

    .

    ±



    x

    =

    y

    -

    4

    Add

    4

    to both sides

    .

    4

    ±



    x

    =

    y

    This is not a function as written. We need to examine the restrictions on the domain of the original function to determine the inverse. Since we reversed the roles of x and y for the original f (x), we looked at the domain: the values x could assume. When we reversed the roles of x and y, this gave us the values y could assume. For this function, / displaystyle x/ge 4x≥4, so for the inverse, we should have / displaystyle y/ge 4y≥4, which is what our inverse function gives.

    The domain of the original function was restricted to / displaystyle x/ge 4x≥4, so the outputs of the inverse need to be the same, / displaystyle f/left (x/right) / ge 4f (x) ≥4, and we must use the + case:

    /displaystyle {f}^{-1}/left (x/right) = 4+/sqrt{x}f

    -1

    (x) = 4+√

    x





    The domain of the original function was restricted to / displaystyle x/le 4x≤4, so the outputs of the inverse need to be the same, / displaystyle f/left (x/right) / le 4f (x) ≤4, and we must use the - case:

    /displaystyle {f}^{-1}/left (x/right) = 4-/sqrt{x}f

    -1

    (x) = 4-√

    x
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