TD

They're asking for the possible values of "5x+2", not of x. So when you say 0 is a possibility, this doesn't mean "x = 0" but "5x+2 = 0". C is the correct answer.

The problem isn't the 4x, but the square! Since you can easily integrate sin(x), it's no problem to integrate sin(ax) with a an arbitrary constant either!

To get rid of the square, use another trig identity:

[math]\cos \left( {2x} \right) = \cos ^2 x - \sin ^2 x = 1 - 2\sin ^2 x \Leftrightarrow \sin ^2 x = \frac{{1 - \cos \left( {2x} \right)}}{2}[/math]

Now you lost the square, went to a cosine and doubled the angle - but that shouldn't be a problem

I can't check your answer since the question isn't 100% clear to me. Is B located 6 miles under the road, vertically? So the distance between B and the road is 6. And the 10 miles, is that the (straight) distance between A and B or between A and the point on the road which is above B?

Ok

Well in general for equations like this, we have that

[math]\cos \alpha = \cos \beta \Leftrightarrow \alpha = \beta + 2k\pi \vee \alpha = - \beta + 2k\pi ,\forall k \in \mathbb{Z}[/math]

I don't consider this as "intuitive" since this holds in general and is a perfect algebraic solution - but perhaps the topic starter can clarify what he meant.

Perhaps it would help if rakuenso would edit out his typo. I don´t really know what the question is about but it´s most certainly not about "cos(x) = cos(x + pi/2)" since disproving this one is trivial: Let x=0 => cos(x)=cos(0)=1' date=' cos(x+pi/2)=cos(pi/2)=0, 1 != 0.

 

Generally, relations involving cosines and sines often become trivial if one rewrites then as the linear combination of exp(ix) and exp(-ix).[/quote']

I think the proposed problem was an equation ("solve for x") and not an identity ("proof for all x").

as stated' date=' is there a none intuitive way of solving

 

[math']

 

cos(x)=cos(x+pi/2)

 

[/math]

 

I know from experience that 7pi/4 and 3pi/4 are the solutions, but is there a way to derive it?

What do you mean with non-intuive way? Two cosines are the same when either the arguments are the same or opposites, with 2k*pi of course.

You're on the right track. With your drawing, split the problem in 2 parts using an intermediate point P. Express the distance as a function of the distance to P by using the pyth. theorem on the triangle, compute the time necessary to do both parts with the given speeds and then add those two parts.

Sorry for any confusion caused by my screwed up awsner.

There's no need to apologize, but do you see what your mistake was?

That's quite fundamental, in general (a+b)^n isn't a^n + b^n.

The answer is indeed 1/27 but be careful, b = 3, not -3 in this case (you substituted correctly though).

I'll go into a bit more detail.

We would like to get rid off that cube root to simplify the numerator. For square roots, we can use the fact that [math]a^2 - b^2 = \left( {a - b} \right)\left( {a + b} \right)[/math] but for cube roots, that won't help.

As I said, we'll be using the identity [math]\left( {a - b} \right)\left( {a^2 + ab + b^2 } \right) = a^3 - b^3[/math]. We consider the current numerator as the factor (a-b) with of course [math]a = \left( {x + 27} \right)^{1/3}[/math] and [math]b = 3[/math]. Now we multiply numerator and denominator with the same factor, namely the second one of our identity, so (a²+ab+b²) with our a and b.

After doing that, instead of cancelling these equal factors (then we wouldn't have done a thing...) we can simplify the numerator since the expression there is now equal to a³-b³ according to our identity. But with our a and b, that becomes [math]a^3 - b^3 \Rightarrow \left( {\left( {x + 27} \right)^{1/3} } \right)^3 - 3^3 = x + 27 - 27 = x[/math] and the cube root is gone, just as we wished.

The only thing that's left is an x, but that can be cancelled out with the x in the denominator leaving only our added expression in the denominator. It's now possible to simply fill in x = 0 in our limit to find the value.

[math](x+27)^{\frac{1}{3}}=x^{\frac{1}{3}}+27^{\frac{1}{3}}=x^{\frac{1}{3}}+3[/math]

Then...

[math]\frac{x^{\frac{1}{3}}}{x}[/math]

So...

As [math]x \to \infty' date=' f(x) \to 0[/math']

And as [math]x \to 0, f(x) \to \infty[/math]

Eeks!

[math]\left( {a + b} \right)^{1/3} \ne a^{1/3} + b^{1/3} [/math] !!!

Use the fact that [math]\left( {a - b} \right)\left( {a^2 + ab + b^2 } \right) = a^3 - b^3 [/math] and consider the nominator as the factor (a-b) in that expression. Then find out what a and b are and multiply nominator & denominator by the second factor in that expression. The nominator now simplifies to a³-b³ and you'll be able to cancel out an x which will get rid off the indeterminate form. Then it's just filling in

I basically said it all, now it's up to you to do the numbers!

That's true but I don't see the relevance of your complex example. Also in the reals, a limit isn't defined if it depends on the way you approach it (meaning: it has to be the same, coming from the left or from the right). In that way, the limit for x going to 0 of 1/x isn't defined because depending on coming from the left or right, you get - or + infinity.

Correct, with that nuance that it could be negative infinity as well of course -depending on the sign of the numerator and the way of approaching 0.

nothing * infinity still equals nothing.

 

0*(any number finite or not) = 0

That's not correct. 0*(a finite number) is 0 but 0*inf is an indeterminate form, just as 0/0, inf/inf, inf-inf, ...

A finite number divided by nothing (zero) is infinity.

Correct.

...

These remarks are correct' date=' up to some imprecise formulation. But I can accept them. [/quote']

Perhaps in the chemistry-forum, the formulation isn't that important but in mathematics, a finite number divided by 0 is not defined.

 

I don't think it's a good idea to confirm his wrong idea about this, certainly not when there was already pointed out that this is wrong and that you have to talk about a limit in order to be mathematically correct.

L'Hopital's rule can only be applied to the cases of 0/0 and inf/inf, other similar forms (such as 0*inf etc) have to be altered to one of the first two forms!

Suppose that it's not necessarily so and we imagine your example but with vectors Xi being linearily dependent in V. Then a certain Xi can be written as a linear combination of the others:

[math]x_i = \sum\limits_{\scriptstyle j = 1 \hfill \atop

\scriptstyle j \ne i \hfill}^r {\alpha _j x_j } [/math]

so then

[math]T\left( {x_i } \right) = T\left( {\sum\limits_{\scriptstyle j = 1 \hfill \atop

\scriptstyle j \ne i \hfill}^r {\alpha _j x_j } } \right)[/math]

but since T is lineair, we then have

[math]T\left( {x_i } \right) = T\left( {\sum\limits_{\scriptstyle j = 1 \hfill \atop

\scriptstyle j \ne i \hfill}^r {\alpha _j x_j } } \right) = \sum\limits_{\scriptstyle j = 1 \hfill \atop

\scriptstyle j \ne i \hfill}^r {\alpha _j T\left( {x_j } \right)} [/math]

and we have written a certain T(Xi) as a linear combination of the other images making them linearly dependent and that's not what we supposed at the start.

I hope I didn't make any silly mistakes

And this is exactly why [math]\frac{d}{dx}\frac{1}{x} = ln(|x|)[/math] and not [math']ln(x)[/math]

Eh... [math]\frac{d}{dx}\frac{1}{x} = \frac{-1}{x^2}[/math].

[math]\int\frac{1}{x} dx = \ln|x|[/math]

Sorry, but that's nonsense.

An infinite series is per definition a sum of an infinite number of terms. The fact that there's no real infinity in 'nature' doesn't mean it doesn't exist in mathematics (since it does, and it only does so because we defined it somehow).

Infinite series surely exist and are very commonly used.

For example, we can define the number e as [math]\sum\limits_{n = 0}^\infty {\frac{1}{{n!}}} [/math]. If it's impossible to determine this infinite sum, then e wouldn't "exist".

You have probably once learned about rational and irrational numbers (which form the real numbers, together). A number was irrational if it has an infinite number of non-repeating decimals. So they never become periodic and they're not finite. It is also impossible to express these number as a fraction of two integers.

Rationals on the other hand, are the numbers which either have a finite number of decimals or which become periodic in their decimals. For example, 22/7 (which is an approximation for pi btw, pi being irrational itself) has the decimal expansion 3.142857142857142857..., or [math]{\rm{3}}{\rm{.}}\overline {{\rm{142857}}} = 3.142857...[/math]. Those last two numbers are merely other notations for 22/7 and its the same number 4 times, a rational one.

I assume it's clear that 0.999... has a repeating part as well (being just 9) and therefore, it has to be rational and thus it can be written as a fraction of integers. You tell me how that can be done, if 1 (=1/1=2/2 or whatever) isn't it.

You're welcome

In that case you already calculated the 2nd derivative so the 'advantage' isn't that big. You are correct though, a zero second derivative and constant sign of the first derivative is a point of inflection.

The reason I asked is that some people apply your method (looking at the sign of the 1st derivative to see if it's constant) to points where the first derivative is zero. Although this gives points of inflection, it is a sufficient but not a necessary condition so you may not find all points of inflection.

The method that I tend to use is' date=' instead of using the double derivative, just use the single derivative for the x value before and after the point which is assumed the point of inflexion. If the sign remains constant, you know the the point is a pont of inflexion.

 

Are there any advantages of using the double derivative instead of the single derivative?[/quote']

You say "the point which is assumed the point of inflexion.", but how did you find this point? With the first or second derivative being 0?

Indeed.

The book had stated that if the double differential was equal to 0, then, it was a point of inflexion. However, my teacher stated that this does not apply on some curves, and therefore, not to use it, but find the gradient before and after the point, and decide thereon.

The fact that the second derivative is zero is a necessary but not a sufficient condition to have a point of inflection. A sufficient condition would be that the second derivative is 0 at the point and changes sign arround the point.