For the function y = f ( x ) = 6/ (2 + x^ 2 ),

Question

a). Find the slope of the line tangent to its inverse function at the point 

I got -2/3 for the slope of f(x) so I thought the inverse was -3/2 but that's wrong.

b).  Find the equation of the line tangent to the inverse function at the point 

Answer 1

To find the inverse of a function simply switch the x and y.

So for the above function.

y = 6 / (2 + x²)

Change it to...

x = 6 / (2 + y²)

Then solve for y by isolating it.

x * (2 + y²) = 6

2 + y² = 6 / x

y² = (6 / x) - 2

y = √( (6 / x) - 2)

Now you take the derivative at x = 2, y = 1

Remember the chain rule.

dy/dx = [1 / (√( (6 / x) - 2) * -2 ] * (6 / x²)

dy/dx = -3 / (√( (6 / x) - 2) * x²)

Substituting x = 2

slope = -3 / (√( (6 / 2) - 2) * 2²)

= 3 / √1 * 4

= -3/4

The reason why you got -3/2 is because you assumed the inverse of the function is f(x) ^-1

^{or 1/f(x)}

When the inverse is reversing the x and y values essentially rotating the function along the y=x line to think visually.

Remember to find the inverse of a function reverse the x and y value.

So the inverse of y = x² Becomes y = √x for x = 0 through infinity

Answer 2

I'm having trouble with this one also, but I can give you some help.

Your answer for the derivative of f is correct.

I cannot get the derivative of the inverse function by formula.(I'll try to think about this some more and get back to you)

So....

To find the inverse function interchange the roles of x and y and solve for y; you should get

y=sqrt[(6-2x)/x] but for x>0 since the original function is not 1-to-1

The derivative of this function is -3sqrt(x)/[x²sqrt(6-2x)] which has value -3/4 at (2,1).

The equation of the tangent then is -3/4=(y-1)/(x-2).

I graphed this on DESMOS and it looks right.

The formula for the derivative of f^-1(x) is 1/f'[(f^-1(x))]...and both you and I forgot to use the correct value in the derivative in the denominator. f'(1)=-4/3 and the derivative of the inverse is therefore -3/4 as I calculated directly from the derivative of the inverse. I hope that helps some more.