# 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).

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

## This Post Has 4 Comments

1. clairajogriggsk says:

Step-by-step explanation:

Check attachment for solution

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

2. itssergioa says:

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. trevorhenyan51 says:

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

$(f^{-1})(2)=-3$

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

4. Fatimaneedhelp says:

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