Use the function f and the given real number a to find (f ^-1)'(a).

Use the function f and the given real number a to find (f ^-1)'(a).


[tex]Use the function f and the given real number a to find (f ^-1)'(a).[/tex]

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  1. Step-by-step explanation:

    Check attachment for solution

    [tex]Verify that f has an inverse. Then use the function f and the given real number a to find (f −1)'(a)[/tex]
    [tex]Verify that f has an inverse. Then use the function f and the given real number a to find (f −1)'(a)[/tex]
    [tex]Verify that f has an inverse. Then use the function f and the given real number a to find (f −1)'(a)[/tex]

  2. 6.

    Step-by-step explanation:

    f(x) = √(x - 4)      x > 4

    y = √((x - 4)

    Square both sides

    y^2 = x - 4

    x = y^2 - 4

    So the inverse:

    f-1(x) = x^2 + 4

    f(-1)'(x) = 2x

    When x = 3

    f(-1)'(3) = 2(3) = 6.

  3. The function

    ... f(x) = (x+2)/(x-1) = 1 + 3/(x-1)

    is symmetrical about the line y=x, hence is its own inverse.

    We can evaluate the desired derivative directly.

    ... f'(x) -3/(x-1)²

    so f'(2) is

    ... f'(2) = -3/(2-1)²

    [tex](f^{-1})(2)=-3[/tex]

    [tex]Erify that f has an inverse. then use the function f and the given real number a to find (f −1)'(a).[/tex]

  4. We have that

    f(x) = x + 2 x − 6 > f(x) = 3x− 6
    The function is continuous in the interval (-∞,∞)

    y=3x-6 >3x=y+6> x=(y+6)/3

    Interchange x and y to find the inverse function
    (f −1)(x)=(x+6)/3> (x/3)+2
    Apply derivative with respect to x
    (f −1)'(x)=1/3 
    for all values of x  (f −1)'(x)=1/3 

    therefore
    for a=3  (f −1)'(a)=1/3

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