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]

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

##
This Post Has 4 Comments

### Leave a Reply

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]

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]

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.

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]

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