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From: Tim Little on 6 Jun 2010 21:14
On 2010-06-07, Albert <albert.xtheunknown0(a)gmail.com> wrote:
> 48x(4x^2 - 1) > 40x
> 48(4x^2 - 1) > 40

You're implicitly assuming here that x > 0.


- Tim
From: Mike Terry on 6 Jun 2010 21:22
"Albert" <albert.xtheunknown0(a)gmail.com> wrote in message
news:38b82ee4-edb3-4afb-a9f1-edc33b41fda3(a)g1g2000pro.googlegroups.com...
> f(x) = (4x^2 - 1)^6, g(x) = (4x^2 - 1)^5
>
> Find {x: f'(x)> g'(x)}
>
> f'(x) = 48x(4x^2 - 1)^5
> g'(x) = 40x(4x^2 - 1)^4
>
> Solve 48x(4x^2 - 1)^5 > 40x(4x^2 - 1)^4
> 48x(4x^2 - 1) > 40x

you have divided both sides by (4x^2 - 1)^4, and this works ok provided
(4x^2 - 1)^4 >= 0. (Which we know holds, so this is OK...)


> 48(4x^2 - 1) > 40

Now you've divided both sides by x, but if x<0 then the inequality needs to
be reversed. E.g. consider

3 * (-1) < 2 * (-1)
but 3 > 2

> 6(4x^2 - 1) > 5
> 4x^2 - 1 > 5 / 6
> 4x^2 > 11 / 6
> x^2 > 11 / 24
> x^2 - 11 / 24 > 0
> Solve x^2 - 11 / 24 = 0
> x < (-1/2) * sqrt(11/6), x > (1/2) * sqrt(11/6)
>
> This doesn't match with the book's answer - what am I doing wrong?


From: Daniel Giaimo on 6 Jun 2010 21:32
On 6/6/2010 8:22 PM, Albert wrote:
> f(x) = (4x^2 - 1)^6, g(x) = (4x^2 - 1)^5
>
> Find {x: f'(x)> g'(x)}
>
> f'(x) = 48x(4x^2 - 1)^5
> g'(x) = 40x(4x^2 - 1)^4
>
> Solve 48x(4x^2 - 1)^5> 40x(4x^2 - 1)^4
> 48x(4x^2 - 1)> 40x
> 48(4x^2 - 1)> 40

Your problem is here. You can't divide through by x unless you know it
is positive. If it is negative then dividing through by it flips the
inequality. If it is 0 you can't divide by it at all.

> 6(4x^2 - 1)> 5
> 4x^2 - 1> 5 / 6
> 4x^2> 11 / 6
> x^2> 11 / 24
> x^2 - 11 / 24> 0
> Solve x^2 - 11 / 24 = 0
> x< (-1/2) * sqrt(11/6), x> (1/2) * sqrt(11/6)
>
> This doesn't match with the book's answer - what am I doing wrong?

--
Dan Giaimo
From: jmorriss on 6 Jun 2010 21:40
On Jun 6, 8:22 pm, Albert <albert.xtheunkno...(a)gmail.com> wrote:
> f(x) = (4x^2 - 1)^6, g(x) = (4x^2 - 1)^5
>
> Find {x: f'(x)> g'(x)}
>
> f'(x) = 48x(4x^2 - 1)^5
> g'(x) = 40x(4x^2 - 1)^4
>
> Solve 48x(4x^2 - 1)^5 > 40x(4x^2 - 1)^4
>         48x(4x^2 - 1) > 40x
>          48(4x^2 - 1) > 40
>           6(4x^2 - 1) > 5
>              4x^2 - 1 > 5 / 6
>                  4x^2 > 11 / 6
>                   x^2 > 11 / 24
>         x^2 - 11 / 24 > 0
>   Solve x^2 - 11 / 24 = 0
> x < (-1/2) * sqrt(11/6), x > (1/2) * sqrt(11/6)
>
> This doesn't match with the book's answer - what am I doing wrong?

Solving non-linear inequalities is not as simple as it may look.

Dividing both sides of an inequality by a variable that may be zero,
or negative, is not a valid operation. You've divided by x and by
(4x^2 - 1)^4

The preferred method is:

First, rearrange the inequality into a form with a zero on the RHS,
and factor the LHS without cancelling anything but positive constants
(like the 8 in your case)

In your case, this becomes, I think:

x ( 4 x^2 - 1)^4 (24 x^2 - 11) > 0

Next, find the zeroes of each factor; in this case, 0, 0.5, -0.5,
+0.5(11/6)^.5, and -0.5(11/6)^.5

Then examine the sign of each of these factors around each of these
critical points. This in turn gives the sign of the overall
function. For example, if x is close to zero and <slightly> positive,
then

x is positive

(4 x^2 - 1) ^ 4 is positive (This one is easy: any even power is non-
negative

(the quadratic) is negative

So x > 0 but close to zero is excluded from the solution

Continue this process at, above, and below each critical popint, and
you're done.

Or just graph the thing...
From: Albert on 6 Jun 2010 21:55
"Mike Terry" wrote:
> "Albert" wrote in message
> > f(x) = (4x^2 - 1)^6, g(x) = (4x^2 - 1)^5
>
> > Find {x: f'(x)> g'(x)}
>
> > f'(x) = 48x(4x^2 - 1)^5
> > g'(x) = 40x(4x^2 - 1)^4
>
> > Solve 48x(4x^2 - 1)^5 > 40x(4x^2 - 1)^4
> >         48x(4x^2 - 1) > 40x
>
> you have divided both sides by (4x^2 - 1)^4, and this works ok provided
> (4x^2 - 1)^4 >= 0.  (Which we know holds, so this is OK...)
>
> >          48(4x^2 - 1) > 40
>
> Now you've divided both sides by x, but if x<0 then the inequality needs to
> be reversed.  E.g. consider
>
>             3 * (-1) < 2 * (-1)
>      but    3        > 2

Right, thank you very much. I've calculated the answer correctly now.
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