Help with these two exam questions

[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...

 

[ppyadof]

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 $\int 1 dx - \int \frac{4}{x+4} dx$ ?

 

[frasifrasi]

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

 

[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?

 

[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)

 

[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

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

 

[ppyadof]

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.

 

[ppyadof]

I assume that by (f-1) you mean the derivative of 1/f(x)

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

 

[CompuChip]

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.

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).

 

[ppyadof]

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).

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

 

[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.

 

[Feldoh]

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

Does that help at all?

 

[frasifrasi]

honestly, no.