# Homework Help: Help with these two exam questions

1. Oct 11, 2007

### frasifrasi

I had these two questions in an exam and do not know how to find the answers...

find (f-1)'(3) for f(x) = sqrt (x^3 + x^2 + x + 6)

and

the int of x/(x + 4) dx

If anyone can get me started on these and show which steps I have to take, that would be great...

2. Oct 11, 2007

I've got an idea about the first 1, but not 100% on it. I think that the derivative of a function is 1/the derivative of the inverse function, so just calculate the df/dx for x = 3, then the answer is the reciprocal of the number you get.

For the second question, ever thought about writing it as $\int 1 dx - \int \frac{4}{x+4} dx$ ?

3. Oct 11, 2007

### frasifrasi

Well, to begin with, how did you transform the integral into that? I triend doing y = (x +4), but it didn't work.

4. Oct 11, 2007

### frasifrasi

can you clarify what you meant for 1... You mean I should take the derivitative of that functiona and plug in 3 for x???

5. Oct 11, 2007

### CompuChip

I assume that by (f-1) you mean the derivative of 1/f(x) (because if it were the inverse, the exercise would become quite messy and ugly). You can do this in two ways (three, actually):
a) Write $$1/\sqrt{\cdots} = (\cdots)^{-1/2}$$ and use the familiar rule.
b) Use the quotient rule
c) Use a rule for $$\frac{d(1/f(x))}{dx}$$ if you know it (otherwise you can derive it using b)

6. Oct 11, 2007

### rock.freak667

I don't really understand what is going on in 1 but 2 looks simple

$$\int \frac{x}{x+4} dx$$

take $$\frac{x}{x+4} = \frac{x+4-4}{x+4} = \frac{x+4}{x+4} -\frac{4}{x+4} = 1-\frac{4}{x+4}$$

so then

$$\int \frac{x}{x+4} dx = \int(1- \frac{4}{x+4}) dx$$

7. Oct 11, 2007

I'm not entirely sure about the first part, but I think that if you find the derivative for f, then put 3 into the resulting equation, and then simply do (your answer)^-1 you will get the derivative of the function f^-1 at x = 3.

The second part, all I did was break the integral up. Is the following not true?
$$1 - \frac{4}{x+4} = \frac{x}{x+4}$$
If it is, then the integral of one (with respect to x), must be the integral of the other (w.r.t x). I just transformed the integral by inspection, there isn't anything special going on there.

8. Oct 11, 2007

yer. I'm assuming he means the inverse of the function. It's not exactly clear.

9. Oct 11, 2007

### CompuChip

That is not true.
As a counterexample, consider
$$f(x) = x; g(x) = f(x)^{-1} = \frac{1}{x}$$.
Clearly, $f'(x) = 1$, so by your argument we should have $g'(x) = 1 / f'(x) = 1 / 1 = 1$ for all x.
But actually,
$$g'(x) = - \frac{1}{x^2}$$
which is in general not equal to 1 (in fact, if x is real, it will never hold).

You really need the quotient rule or "power rule" (I don't know what it's called - if it has a name - I mean the rule dx^n/dx = n x^(n-1)) to find the derivative of 1/f(x).

10. Oct 11, 2007

I said that I wasn't sure about it - Thats my disclaimer :P

11. Oct 11, 2007

### frasifrasi

I meant the derivative of the inverse of the function at 3... Thanks for the help with the second question - I can't believe they did that +4 -4 thing!!!!

I would still like help with 1, though. The answer is supposed to be 1.

Last edited: Oct 11, 2007
12. Oct 11, 2007

### Feldoh

$$\frac{d}{dx}f^{-1}(x)=\frac{1}{f^{'}(f^{-1}(x))}$$

Does that help at all?

13. Oct 11, 2007

### frasifrasi

honestly, no.