powers, tangents, parallel lines
Chris
Hi,
I have a homework question which reads:
Find the values of x so that the tangent to the function y=3/(x^(1/3)) is
parallel to the line x + 16y + 3 = 0
here is what I have come up with.
f(x)=3x^(-1/3)
f '(x) = -x^(-4/3)
f '(x) = -1/(x^(4/3))
x+16y+3=0
y = (-1/16)x - 3/16
so the slope of the given line is -1/16, and the slope of the tangent to the
given function is -1/(x^(4/3)
-1/16 = -1/(x^(4/3))
16 = x^(4/3)
x=8
However, my text wants me to have x=-8 as a solution as well as x=8. Trying
to calculate -8^(4/3) gives an error. Is there somewhere in my calculations
that I could have done something different, which would give x=-8 as an
alternative answer?
Thank you.
Chris
Bob Pease
"Chris" <bir...@hotmail.com> wrote in message
news:CpycndJGlMs...@cyg.net...
the fourth root has four answers.
in this problem, two of them are real .
for example x^4 = 1 has
1, -1, i, -i as roots on the complex plane .
in your problem,
(x^4)^(1/3) works in Excel for both answers
, but (x^(4/3)) blows up. with -8
RJ Pease
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Ken Pledger
wrote:
> ....
>
> 16 = x^(4/3)
>
> x=8
>
> However, my text wants me to have x=-8 as a solution as well as x=8. Trying
> to calculate -8^(4/3) gives an error....
The words "gives an error" have a calculator flavour. A calculator
using logarithms to find powers can sometimes run into trouble with
negative numbers. Just pick up a pen.
The power x^(4/3) can be handled either as (x^4)^(1/3) or as
(x^(1/3))^4. Either way, writing x = -8 = -2^3 (with your calculator
safely out of reach) should give you the answer 2^4 = 16.
Ken Pledger.
Barb Knox
> Sorry if this posts twice, strange things happening here...
If you feed (-8)^(1.333333333333) into a real-number calculator it will
indeed choke. But consider that x^(4/3) = (x^(1/3))^4.
16 = (x^(1/3))^4,
so +/- 2 = x^(1/3),
so +/- 8 = x.
There may be a moral in this about relying too heavily on calculators.
--
---------------------------
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Paul Sperry
> Sorry if this posts twice, strange things happening here...
>
>
> Hi,
>
> I have a homework question which reads:
>
> Find the values of x so that the tangent to the function y=3/(x^(1/3)) is
> parallel to the line x + 16y + 3 = 0
>
> here is what I have come up with.
>
> f(x)=3x^(-1/3)
> f '(x) = -x^(-4/3)
> f '(x) = -1/(x^(4/3))
>
> x+16y+3=0
> y = (-1/16)x - 3/16
>
> so the slope of the given line is -1/16, and the slope of the tangent to the
> given function is -1/(x^(4/3)
>
> -1/16 = -1/(x^(4/3))
> 16 = x^(4/3)
So, x = 16^(3/4) = (+/-2)^3 = +/-8. You really have 16^3 = x^4 so
x^4 = 4096 so x = +/-8.
[...]
> However, my text wants me to have x=-8 as a solution as well as x=8. Trying
> to calculate -8^(4/3) gives an error.
This is a calculator peculiarity. If you try either (-8)^(1/3) then take
that result to the fourth power or (-8)^4 then the cube root of that
answer you will get 16.
> Is there somewhere in my calculations
> that I could have done something different, which would give x=-8 as an
> alternative answer?
Except for forgetting that there are two fourth roots in the reals, you
did fine.
--
Paul Sperry
Columbia, SC (USA)
Chris
Thanks everyone, it was indeed a case of trying to calculate before
thinking. Having not learned about logs, or exponentials to any degree
beyond the very basics, I am now interested in why a calculator cannot
evaluate this expression, when it does in fact have a real result? If
anyone could shed light on this, I would appreciate it.
Thanks again!
Chris
Bob Pease
Calculators use well-known infinite series to approximate stuff like logs,
exponentiation , and trig functions.
They do this by storing a temporary partial answer, and then doing things
over and over again until they cantg get any noticeable difference in
succesive partial sums.
I don't know the details, but if they get stuck in a loop, overflow a
register or go past a certain mumber of repeating, they abort and then
report what they got, or give an error message.
for some reason the calculator and Excel gets stuck on this particular
order of operations, by calculating a decimal approximation of 4/3 and
trying to plug that into an infinite series rather than to do two problems
of series , chained up.
.
RJ pease
Doug Magnoli
non-integer power, it goes to pieces. If the power's an integer, it can
calculate it by x*x*x*..., but if it's not an integer, it does the problem
using logs. And there's no real log for a negative number.
Some calclators have cube-root keys, and usually you can feed such a
calculator something like cbrt(-1/8) and it'll tell you -0.5. But if you try
to do the same thing by raising -1/8 to the power of the reciprocal of 3,
you'll get an error, because then it tries to do it by logs.
However, it's far better to think these things through yourself than to rely
on what the calculator's telling you. Calculators, as you've seen, don't
always tell you the whole truth.
I've got a calculator that does complex numbers (RPNCalc, by Jeffry Baker, for
the Mac), and you sparked my curiousity: what happens if I asked it for
(-1/8)^(1/3)? It doesn't give 0.5 as a result--it gives 0.25 + 0.43301 i.
Well, we know that the three roots of -1/8 will be -1/2 * exp(2k i pi/3),
where k = 1, 2, 3. And sure enough, 0.25 + 0.43301 i = 1/2 * e^(2 pi i/3).
(And when I ask the calculator to cube that, it gets -0.125.) So even
calculators that do complex arithmetic can't always be relied on to give you
what you were expecting.
Mathematica, on the other hand, when asked to give (-1/8)^(1/3), gives
1/2*(-1)^(1/3), a pretty complete answer, actually, since (-1)^(1/3) = e^(i pi
(1+2k) / 3), k=0,1,2 (or k=1,2,3), which really is all the cube roots of
-1/8. When I ask it for (-1)^(1/3) numerically, it doesn't say -1, it says
0.5 + 0.866025 i, notice that that's the same thing the complex number
calculator gave.
-Doug Magnoli
[Delete the two and the three for email.]
Paul Sperry
Doug Magnoli <dmagn...@attbi.com> wrote:
[...]
>
> Mathematica, on the other hand, when asked to give (-1/8)^(1/3), gives
> 1/2*(-1)^(1/3)
[...]
Maple decrees "(1/8)*(-1)^(1/3)*8^(2/3)".
Randy Poe
[ (-8)^(4/3) ]
> Thanks everyone, it was indeed a case of trying to calculate before
> thinking. Having not learned about logs, or exponentials to any degree
> beyond the very basics, I am now interested in why a calculator cannot
> evaluate this expression, when it does in fact have a real result?
It has three results, one of which is real and two
of which are complex. Some computer mathematics programs
will return a compex root if asked for just one, some
might return the real root, and some might give errors
as your calculator did because the algorithm involves
taking a logarithm of a negative number.
Calculators typically use floating-point arithmetic and
don't check whether something is an integer and a different
algorithm might work. A calculator doesn't store (4/3)
exactly. It will calculate (-8)^1.33333333. I don't know
if this expression has any real values.
- Randy