Question

f(x)=3/x^2 and g(x)=f^(-1) (x) find the slope of the curve g(x) = f -1(x) at the point (3, 1).

Answer 1

First, let's find the inverse of \(\frac{3}{x^2}\). We have to solve for \(x\).
$$y = \frac{3}{x^2} \\
yx^2 = 3 \\
x^2 = \frac 3 y \\
x = \sqrt{\frac 3 y}$$

Answer 2

This means \(f^{-1}(x) = \sqrt{\frac 3 y}\). Now, can you take the derivative of that function if I write it like this?
$$(3y^{-1})^{1/2}$$

Answer 3

-1/2(sqrt3)(1/y)^3/2

Answer 4

$$(3y^{-1})^{1/2}\\
-\frac 1 2 (3y^{-1})^{-1/2} y^{-2}$$
Either you've got it and I'm reading it wrong or you're very close.

Answer 5

I probably be wrong ha

Answer 6

Alrighty. Well, what I have is true. So now, we just plug the point x = 3, or y = 3 in this case to find the slope at the point (3, 1).

Answer 7

?

Answer 8

Plug y = 3 into \(-\frac 1 2 (3y^{-1})^{-1/2} y^{-2}\) to obtain the answer.

Answer 9

isn't y=1? at point (3,1)?

Answer 10

Well, we solved for x. We can treat \(y = x\) now, because \(g(x)\), the inverse, is a function of \(x\).

Answer 11

Make sense?

Answer 12

-1/6?

Answer 13

Yes.