Math Is Fun Forum

If you use the cubic formula, shouldn't you come up with the exact root?

First, take the equation:

x(x-15) + y(y-10) + 78 <= 0

and complete the square.  You should end up with an equation:

(x-a)^2 + (y-b)^2 <= r^2

where a, b, and r are constants.

Now how much calculus do you know?  Since you are finding min and max's, I have to assume you know some, but this seems like a Multivariable Calculus question.  Is this what you're taking?

If so, all you need to do is derivative with respect to x and y, find the zeros and the end points on the region.  Then find which one of these is the highest and which is the lowest.

So what is the exact root of it, krassi?

I've never heard of it thought that way, but 1 is definitely L and 7 is definitely T.

1'v3 n3v3r h34rd 0f 17 7h0u6h7 7h47 w4y, bu7 1 15 d3f1n173|y | 4nd 7 15 d3f1n173|y 7.

1'\/3 |\|3\/3|2 |-|3@|2[) 0|= 1+ +|-|0|_|6|-|+ +|-|@+ \X/@`/, |}|_|+ 1 1$ [)3|=1|\|1+3|`/ | @|\|[) 7 1$ [)3|=1|\|1+3|`/ +.

No, it can't be factored.  The only real factor is a real number x~1.32.  So any attempts to factor it would be futile.

To put another nail in the coffin...

If you think that 4.0 rounds down to 4, then it is also true that 5.0 rounds up to 5.  So you would have:

4.0 .1 .2 .3 .4 .5 .6 .7 .8 .9 5.0

4.0, .1, .2, .3, and .4 round down

4.5, .6, .7, .8, .9, and 5.0 round up

So there is still 5 numbers that round down, and 6 numbers that round up.

Oy, I completely missed a much more direct way to do the limit:

Since x+2 approaches zero as x approaches -2, x+2 is very close to 0 before it gets there, in other words, very small.  (x+2)^(1/3) then also becomes increasingly small, and multiplying this by 3 has basically no effect as it will also become increasingly small.  So 1 over this means it goes towards infinity.

f(x) = abs(x), a corner exists at f(0).  It is literally a corner.

As for the answer, it is D:

For a verticle tangent, the function's slope must approach infinity as it approaches the point.

So what we want is:

Multiplying this by

Now x+2 approaches 0 as x approaches -2.  But we know that f(x)^(1/3) > f(x) if f(x) < 1.  So the numerator gets (relatively) larger and the denominator gets smaller as x approaches -2.  Therefore the slope approaches infinity, and you have a verticle tangent.

Heh, 30 hours?  Jeez, talk about no patience...

I didn't believe you till I tried it, and you're right, it doesn't work in different styles.  It's a fairly simple task to get the style the person is viewing, then change the font to the background color of the post.  Weird.

So then seerj, why is:

1 3

Not the smallest solution?

Going back on what I wrote before...

Seerj, you keep saying:

Which means that you need a list which contains every number from 1 to n.  If you make a list like this, it has to have size n.  Ignoring the "perfect square" part:

1, 4, 6, 3, 2, 5

That would be a correct list.  However:

1, 4, 2

Would not be a correct list because it doesn't include 3, which is a number from 1 to n, n being 4.

Is this correct?  And if it is, how is:

A correct sequence since it doesn't contain 2, 4, 5, 7.. etc?