Hi all,

Thank you in advance for your help. I'm scratching my head somewhat trying to see just where I went wrong (if I went wrong) on a problem related to inverse functions. The question reads as follows:

Here's the methodology I used, which admittedly may be far off course:Use $\displaystyle

$$g'(x) = {1 \over {f'(g(x))}}$$$ where g is the inverse function for f to compute the requested derivative.

Let $\displaystyle

$$f(x) = {{{x^3}} \over 2} - {2 \over {{x^3}}}$$$ on the interval 0>x . Determine an equation for the line tangent to the graph y=g(x) at $\displaystyle

$$x = {3 \over 2}$$$

First I found f'(x) since I was looking at f(x) already:

$\displaystyle

$$\eqalign{

& f(x) = {{{x^3}} \over 2} - {2 \over {{x^3}}}$

$\displaystyle & f'(x) = {3 \over 2}{x^2} + 6{x^{ - 4}} \cr} $$$

Since f(x) and g(x) are inverse functions, (x,y) for f(x) is (y,x) for g(x). Based upon that, I took the x value I would be looking for in my derivative $\displaystyle

$$x = {3 \over 2}$$$ and put it into my original formula to see what the corresponding g(x) value would be.

$\displaystyle

$$Let\,x = g({3 \over 2})$$$

$\displaystyle

$${3 \over 2} = {{{x^3}} \over 2} - {2 \over {{x^3}}}$$$

$\displaystyle

$${1 \over 2}{x^3} - {2 \over {{x^3}}} - {3 \over 2} = 0$$$

$\displaystyle

$$x = 1$$$

$\displaystyle

$$g({3 \over 2}) = 1$$

$

Since I now know the g(x) value, I used the formula to calculate g'(x):

$\displaystyle

$$g'({3 \over 2}) = {1 \over {f'(1)}}$$ $

$\displaystyle

$$g'({3 \over 2}) = {1 \over {{3 \over 2} + 6}}$$$

$\displaystyle

$$g'({3 \over 2}) = {2 \over {15}}$$ $

The tex got a little crazy on that one line, but hopefully you catch my drift. So with the ordered pair for g(x) of $\displaystyle

$$({3 \over 2},1)$$$ I went back to the point-slope formula of a line:

$\displaystyle

$$y - {y_1} = m(x - {x_1})$$$ where $\displaystyle

$$m = g'({x_1})$$$

I know what g'(x) is, so I use that for my slope and end up with:

$\displaystyle

$$y - 1 = {2 \over 15}(x - {3 \over 2})$$$

Yielding a final solution of: $\displaystyle

$$y = {2 \over {15}}x - {1 \over 5}$$$

Now, when I try and check my work via some calculus utilities, I'm not getting that same answer. I'm getting something else. What I'm basically looking for is verification that I'm doing this correctly, or help in identifying where I'm going wrong.

I really appreciate your time. Thanks!