I'm trying to do some last-minute math revision, as I have a test on Monday, and I found a question I can't do. And the mark-scheme doesn't make sense.

The question:

f:x |-> 3^x

Find the solution of the equation f''(x) = 2

My first step was to find the inverse of f(x)... f'(x)

So I let k = 3^x
ln(k) = ln(3^x)
ln(k) = x(ln(3))
x = (ln(k))/(ln3)
(x = log(base 3)(k))

But then I tried finding the inverse of the inverse... f''(x)and got the original function.

The answer scheme, however, says that f'(x) is not what I got at all, but is actually (3^x)(ln(3)).

What am I doing wrong?

Well I'm not actually sure what you mean by inverse, what are trying to do with it, or why it would help you get f'(x). You are however correct that x = log(base 3)(k), but I would say that should be obvious from the initial problem, as it is basically the definition of a logarithm.

Here is how I would figure out f'(x) and f''(x):
f(x) = 3^x
f(x) = e^(x * ln(3))
f'(x) = d(e^(x * ln(3)))/dx
f'(x) = e^(x * ln(3)) * ln(3)
f'(x) = 3^x * ln(3)
f''(x) = d(3^x * ln(3))/dx
f''(x) = d(e^(x * ln(3)) * ln(3))/dx
f''(x) = ln(3) * d(e^(x * ln(3)))/dx
f''(x) = ln(3) * ln(3) * 3^x

[ 06-04-2005, 01:58 AM: Message edited by: Seraph ]

Dammit! That's where the problem lies! I knew I was forgetting something... f'(x) doesn't mean the inverse of x, it means the derivative of it!

D'oh!

Thanks! :D

[ 06-04-2005, 02:04 AM: Message edited by: Callum ]

and f''(x) is the second derivative

f-1(x) is the inverse i think.

Yeah... bodes well for monday's exam no? :D

the inverse is 1/f(x) and onviously, doing 2 inverses would be 1/1/f(x) which =f(x), hence it got Callum nowhere [img]smile.gif[/img]

[ 06-04-2005, 08:21 AM: Message edited by: Aragorn1 ]

Please tell me this is not Grade 10 math!! :'(

my math exam has been very easy so far, 25 minutes for 15 multiple choice questions [img]smile.gif[/img]

I'm in the 'year 13' although we wouldn't call it that. I'm at college, so i'm a second year. I think it used to be called 'upper-sixth,' but only the oublic schools call it that now i think.

Maths multiple choice!?

*blinks*

Wow.. Sounds... difficult.

And I'm in the year below Aragorn1... though we do call it year 12/13 here.

Does your school go from yr 7-13?

...

o.o

...

*brain explodes* I envy people that understand that crap [img]tongue.gif[/img]

...

o.o

...

*brain explodes* I envy people that understand that crap [img]tongue.gif[/img] </font>[/QUOTE]No you don't.

And these days that's the 'easy' stuff for me! :D
Double angle formulae, trigometric integration AHHHH!

Oh man, I do math at university level, and someone mistook the inverse for the derivative! :D :D :D :D

No offense, there are people who don't know what the '2' above the x means. ;)

I don't know if anyone is interested, but my girlfriend was able to come up with a totally different way to get f'(x).

f(x) = 3^x
ln(f(x)) = ln(3^x)
ln(f(x)) = x*ln(3)
d(ln(f(x)))/dx = d(x*ln(3))/dx
d(ln(f(x))/df(x) * d(f(x))/dx = d(x*ln(3))/dx
1/f(x) * f'(x) = ln(3)
f'(x) = f(x) * ln(3)
f'(x) = 3^x * ln(3)
You should then be able to apply the same procedue to get f''(x).

...

o.o

...

*brain explodes* I envy people that understand that crap [img]tongue.gif[/img] </font>[/QUOTE]I'm with you, but I don't envy them. Anything with x's are too much for my brain to handle. [img]tongue.gif[/img]

<font color = lightgreen>Seraph, your girlfriend got the right answer and differentiated the function correctly.

Then function is question is an exponential function; in its generic form looks like this: f(x) = k ^ x, with k being a non-negative numbers (usually). The first derivative takes the form f'(x) = (ln k) * (k ^ x). Multiple derivatives have this form: (f(n))(x) = ((ln k)^n) * (k ^ x).

If it helps, you could always think of letting k = e, because finding derivatvies for e^x is really easy. [img]graemlins/beigesmilewinkgrin.gif[/img]

Interestingly, if you let k = sqrt(2), begin with an x-value of 1, and iterate the function (the x-value of the next step is the f(x)-value from the last step) then the iterated function values converge to sqrt(2). [img]graemlins/petard.gif[/img] </font>

2*x isn't it? :D :D :D

Does your school go from yr 7-13? </font>[/QUOTE]Yes. :D

...

o.o

...

*brain explodes* I envy people that understand that crap [img]tongue.gif[/img] </font>[/QUOTE]No you don't.

And these days that's the 'easy' stuff for me! :D
Double angle formulae, trigometric integration AHHHH! </font>[/QUOTE]Pfft... I know double angle formula.

And that should be easy stuff for me too... and is, now that I revised. My math exam went well... except for a right cow of a last question, involving 3d calculus and vectors and god knows what else. He said we wouldn't be having anything on planes and lines in 3d space... then gave it us anyway. [img]tongue.gif[/img]

Yep... I know that know... yay for my extra three hours of revision! :D

What exam was it? Internal, external?

Internal... but my predicted grades will kinda hinge on this. [img]tongue.gif[/img]

Wow, that's a bit tough! In my case it was 'meh, you'll probably get an A'. God knows what ot would have been if they went on my maths test scores. Good for uni applications! THen again, i've done the ASs, so that would give them a better indication.

I hate you all now [img]tongue.gif[/img] :D