Precalculus Example 4.1.4(3) - Suspected very subtle error

classic Classic list List threaded Threaded
4 messages Options
Reply | Threaded
Open this post in threaded view
|

Precalculus Example 4.1.4(3) - Suspected very subtle error

MarkAU
This post was updated on .
I'm a retiree working my way through the Precalculus text to keep my brain exercised and to settle a score with myself for the maths that I didn't persist with through to the end of high school.

Carl and Jeff, thank you for making your great textbooks freely available to all.

I'm halfway through Precalculus, and while taking a revision break and revisiting previous material, I found a suspected very subtle error in Example 4.1.4(3) on page 309.

I don't have a graphing calculator (I rely on an HP 35s Scientific), so I use polynomial division to ascertain whether the graph of a function is approaching an asymptote from above or below.

In Example 4.1.4(3), the graphing calculator appears to indicate that as x -> -infinity, h(x) approaches the asymptote y = -3 from above; and that as x -> +infinity, h(x) approaches the asymptote y = -3 from below.

By polynomial division, 6x^3 - 3x + 1 divided by -2x^3 + 5 yields a quotient of -3 (the horizontal asymptote) and a remainder of -3x + 16, or:

p(x) / d(x) = -3 + ((-3x + 16) / (-2x^3 + 5))

The fraction ((-3x + 16) / (-2x^3 + 5)) will determine whether the graph is approaching the asymptote from above or below.  If x -> -infinity, the numerator and denominator will both be positive in value, and the fraction will approach zero as an increasingly small positive value.

However, if x -> +infinity, the numerator and denominator will both be negative in value, and the fraction will in this case as well approach zero as an increasingly small positive value, ie, the graph will approach the asymptote y = -3 from ABOVE.

That is not what the graphing calculator appeared to predict, nor consistent with your stated solution to the problem, and I was figuratively banging my head against the wall wondering where I went wrong.  Then I did some calculations and found:

* h(5) = 736 / -245 ~= -3.0041 (which is an unsurprising result)

* h(6) = 1279 / -427 ~= -2.9953 (ahaa - the graph has crossed the asymptote!)

[EDIT: After further thought, I could have saved myself some grief if I had simply solved (-3x + 16) / (-2x^3 + 5) = 0 and determined that the graph would hit the asymptote at x = 16 / 3]

Pressing on and narrowing calculations down to increments of 0.1, a local maximum appears to be reached at around h(8.0) ~= -2.992149.  h(100) ~= -2.999858, and I didn't go higher in order not to exhaust digits of accuracy available on my calculator, but that result is consistent with the prediction from my polynomial division that as x -> +infinity, h(x) approaches the asymptote y = -3 from above.

You'd have to zoom the graphing calculator in almost microscopically to see that (if you were lucky!).

Do you agree?
Reply | Threaded
Open this post in threaded view
|

Re: Precalculus Example 4.1.4(3) - Suspected very subtle error

Carl Stitz
Administrator
Hello!

Looks like you've discovered some "hidden" behavior!

Nicely done!  Indeed there is a local maximum!

I'll pass this along to Jeff and we can fashion this into a nice extension exercise.

Thanks for reaching out!

Carl

Reply | Threaded
Open this post in threaded view
|

Re: Precalculus Example 4.1.4(3) - Suspected very subtle error

MarkAU
Carl, you can annoy Jeff by pointing out that this instance illustrates the pitfalls of relying for your evidence on the "low hanging fruit" approach of easy resort to technology.
Reply | Threaded
Open this post in threaded view
|

Re: Precalculus Example 4.1.4(3) - Suspected very subtle error

Carl Stitz
Administrator
Will do!