1. ## Inverse function exercises

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

Let $
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 $
x = {3 \over 2}$
Here's the methodology I used, which admittedly may be far off course:
First I found f'(x) since I was looking at f(x) already:


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

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

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

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

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

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

$
x = 1$

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

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

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

$

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

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

$
y - {y_1} = m(x - {x_1})$
where $
m = g'({x_1})$

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

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

Yielding a final solution of: $
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!

2. Hi,

In the beginning when you set your function f(x) = to 3/2, and solved it you got x = 1. When I did it, I got x = -1, and as a result g(3/2) = -1, and my tangent line came out to be y = -2/9(x - 3/2) - 1.

3. Hi,

In the beginning when you set your function f(x) = to 3/2, and solved it you got x = 1. When I did it, I got x = -1, and as a result g(3/2) = -1, and my tangent line came out to be y = -2/9(x - 3/2) - 1.
Ah I left out one key detail in the problem description...the problem is over the interval 0<x .

Edit: Although you're right...that's where I went wrong. -1 is a valid answer, however the other answer is x=1.587, not x=1. That's where I made my mistake...I wrote x=1 instead of x=1.587 . Now it adds up! Thanks!

At least the methodology was right. That was my primary concern.

4. Originally Posted by Malaclypse
Ah I left out one key detail in the problem description...the problem is over the interval 0<x .
But in the original problem it said 0>x (your first post). Was that incorrect?

5. Apparently I DID mention the interval, but I got that wrong too.

Yes it was wrong. It was supposed to be 0<x. My brain is apparently a bit fried this evening.

6. Well glad that you got it resolved. Yeah, your methodology was completely correct.