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That's a fair point actually, it does, I just wanted a value of *x* that would make my algebra nice and easy, so let's try something else:

I believe it comes to the same thing, doesn't it?

**Au101**- Replies: 3

Okay, this is the last question on this exercise and it's got me completely confused in so many ways. This is the question:

Now, I've got all of the pieces of the puzzle. Obviously:

Next, I said:

Now, if we sub that value of *x* back into *y*, we get this:

So we've got the two parts of our inequality, but I'm just stuck "considering the minimum value of the function". I tried finding the second derivative:

Now, if I put *a*/2 into that, I get:

which I reckon is surely positive if *a* > 0, which suggests that *a*/2 is our minimum. The bad news is that Wolfram|Alpha thinks that the function has no global minimum. The worse news is that I actually found *y* to be a maximum at *x* = *a*/2 when I tried this approach:

Again Wolfram|Alpha thinks that the function has no global maximum, either.

So I'm a little bit at sea here - I'm clearly doing something wrong, I just don't know what!

I was wondering if anybody could suggest a starting point for the second proof, just to get me going

Perfect! thank you!

**Au101**- Replies: 5

Another tricky little question :S

I wasn't really sure where I was going with this so I decided to just dive right in and differentiate:

As for the second part of the first proof, I'm pretty stumped. We know that *k* > 0, so for *k* < 1, *k* - 1 will be negative and an element of the set of real numbers between 0 and 1 (exclusive). In other words:

So you will have a fraction and the exponent of *x* will not be an integer. Whereas, for *k* > 1 you will have a positive exponent, which may be an integer and may, in fact, be any positive real number. That's the closest thing to a brainwave I've had really - do you have any ideas?

Thanks Olinguito, that's the sort of thing I would have expected, but the exercise in question occurs in the book before implicit differentiation is introduced, which is why I was hesitant to go down that kind of road.

Perhaps the book was expecting you to cheat, or perhaps I've simply got lucky.

Either way, I appreciate it and I'm more or less happy!

Edit: Although, surely

Is not *V* but *V*²? So I should have

**Au101**- Replies: 3

Another quick question.

The question I've just completed reads:

I completed the proof and then got a little bit stuck on the second half of the question. What I was minded to do was square root both sides in order to obtain an expression in terms of plain old *V* and then use the chain rule. This turned out to be a bit problematic and, in the end, I had to resort to what I thought was a bit of a cheat, but it was the only way I could get the right answer. Instead, I did:

Solved

for *x*:

Found that

And solved for *V*: *V* = 36.

Question: Is this legitimate? More to the point, is it generally legitimate, cause I've got things like implicit differentiation in the back of my mind and something feels wrong about differentiating *V*² but I couldn't think how else to do it!

Thank you very much for your help

(This was just a part of a question in a differentiation exercise)

**Au101**- Replies: 2

Just a quick one to ask what is the best way to solve the following equation:

My approach was to say:

And then square both sides:

However, -3/4 obviously isn't a solution to:

If you take the principal value of the square root function and it is also not given as a solution by wolfram alpha.

So I was wondering if I went about this question the right way, or should I have done something different?

Thank you very much Olinguito, that's got it, I just needed help with how to think about it

The final question of this exercise is (in my opinion) another slightly fiendish one. Again, there's something I'm just not getting. Here it is:

For what it's worth, I've worked out that the gradient (call it *m*) of the line is:

But I can't say I've had any more breakthroughs than that.

I wondered briefly whether the least value of *OA* + *OB* will occur when the gradient is -1. If that's true I haven't been able to prove it, or get anywhere by taking that as a starting point!

Yeah - well, certainly, the answer checks out (26 when rounded appropriately).

Thanks a lot! That seems to've done the trick

I was about halfway there, but I hadn't quite fully understood the question.

**Au101**- Replies: 7

A fun little question which I'm afraid has got me stumped. I'm not really sure how to approach this, what do you think?

Sorry Olinguito, I didn't realise you'd posted, but we're both basically saying the same thing!

Hi

The " is not a symbol that LaTeX recognises. More importantly, you should have a space between your \cos and your A, because it reads \cosA all as one string and \cosA is not a command it recognises - that's why you're getting an error. To be honest, you only really needed

h = c\cos A

is it important for you to have quotation marks around your \cos A. If it is you could do that, but I'm not sure I really understand why you would want to.

Also, did you mean to get *Delta* or were you hoping for Δ (\Delta)?

bob bundy wrote:

That's why I suggested a simple example first. When I find a bit of maths hard, I look for simple cases first so I can slide into the hard stuff in small steps.

Try it with simple functions and just two points, letting the second move gradually closer to the first. If you set up the formulas on a spreadsheet it will be easy to change 1.1 into 1.01 into 1.001 ...........

Then change the fixed pint t=1 into t=2 and repeat. Or change the formulas into harder ones.

Think I'd better sign off now and wash away the literal garden mud and metaphorical middle lane hogging driver anguish. Good night. See you tomorrow.

Bob

Sleep well, and thanks for delaying your well-earned sleep for me!

*f(t)* nor *g(t)* is invertible, though? Surely then we could not express *t* as a function of *x* or as a function of *y* and then we could not express *y* as a function of *x* by this method?

bob bundy wrote:

With my simple functions

x = 2t so t = (x/2)

Therefore y = (x/2)^2

It would be somewhat harder with the functions you chose.

Bob

Yes I see that, so we can see that since both *x* and *y* are functions of *t*, there will be some way of relating *x* and *y*. They are both expressed in terms of *t* so what we can do is express *t* as some new function of *x* and thereby express *y* in terms of *x* (since *y* is a function of *t* and we have just expressed *t* as a function of *x*, so *y* may also be expressed as a function of *x*.) Clearly, in some cases *y* will have to be expressed as a function of *x* and *t*.

I see - I think, but I feel like I'm missing something deeper. It's like I'm almost there with it, but not quite. I can see that where you have the points:

You can say that the gradient (say *m*) is given by the change in *y* over the change in *x*. That is to say:

And that *δx* and *δy* must both be functions of *t*, but I guess I'm not quite sure I see why the gradient of the graph *y = φ(x)* is equal to gradient of the graph *y = g(t)* divided by the gradient of the graph *x = f(t)*. It feels kind of right and yet, I don't see it somehow.

Edit: obviously that should really be the gradient of the curves at an equivalent point - it's actually not very easy to express!

*y* may be expressed as a function of *x*, although quite what that function would be I don't know - a function in terms of *t*, it seems.

bob bundy wrote:

but I wouldn't want you to stay up on my account!

I've just driven 70 miles round the M25 so my brain is wide awake, even though I'm theoretically tired having spent the day doing jobs for my Mum. So this is probably a good way to unwind.

Bob

Ooooh I have mostly bad memories of the M25