[SOLVED] Piecewise functions

dwsmith

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Mar 2010
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If \(\displaystyle f\) is the function defined by
\(\displaystyle f(x) =
\begin{cases}
xe^{-x^2-x^{-2}}, \mbox{if x}\ne 0 \\
0, \mbox{if x}=0
\end{cases}
\)
at how many values of x does the graph \(\displaystyle f\) have a horizontal tangent line?

What is a good way to approach this problem?
 
May 2010
16
6
Sydney - Australia
hello dwsmith,

is this a general question on how to find the turning points?

if it is, i would first find the derivate of the function, then plug in zero in there...those values would be the x-coordinate i will need to keep, then i take those x-coordinates and plug 'em into the first function...that'll be your y-coordinate. Then how many of those point you tend to have depending on the question, you state them as your horizontal tangents.

i hope thats what you were asking for :)

-sugarT
 
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dwsmith

MHF Hall of Honor
Mar 2010
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582
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hello dwsmith,

is this a general question on how to find the turning points?

if it is, i would first find the derivate of the function, then plug in zero in there...those values would be the x-coordinate i will need to keep, then i take those x-coordinates and plug 'em into the first function...that'll be your y-coordinate. Then how many of those point you tend to have depending on the question, you state them as your horizontal tangents.

i hope thats what you were asking for :)

-sugarT
I don't think it is that simple; plus, the first derivative test identities maxs and mins.
 
Apr 2010
384
153
Canada
I don't think it is that simple; plus, the first derivative test identities maxs and mins.
Why not? The derivative is the slope of the tangent, and if we set that derivative (i.e. the slope of the tangent) equal to 0 this identifies all places where there is a horizontal tangent.

I think you may be getting confused with local max/min and overall max/min. Equating the derivative equal to 0 finds all of these! Not just the overall. We actually become acustom to thinking this way because most of the problems we compute deal with absolute max/min but we forget, we can be dealing with a local max/min.

I also think the real "hard part" to this problem is finding our x's when we equate this to 0. I mean, it looks like there might be something very messy to work with after we take the derivative (though this could be partially because im half sauced at the moment...)
 
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dwsmith

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\(\displaystyle \frac{d}{dx}\left [xe^{-x^2-x^{-2}}\right]=e^{-(x^2+x^{-2})}(3-2x^2)=0\)

\(\displaystyle x=\pm\frac{\sqrt{6}}{2}\)

\(\displaystyle \frac{1}{e^{x^2+x^{-2}}}=\frac{1}{e^{x^2}e^{x^{-2}}}=1\neq 0\)

Therefore, I have 2 answers but there are 3. What is the 3rd answers?
 

undefined

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Mar 2010
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Chicago
\(\displaystyle \frac{d}{dx}\left [xe^{-x^2-x^{-2}}\right]=e^{-(x^2+x^{-2})}(3-2x^2)=0\)

\(\displaystyle x=\pm\frac{\sqrt{6}}{2}\)

\(\displaystyle \frac{1}{e^{x^2+x^{-2}}}=\frac{1}{e^{x^2}e^{x^{-2}}}=1\neq 0\)

Therefore, I have 2 answers but there are 3. What is the 3rd answers?
Hmm, the derivative I get is

\(\displaystyle e^{-\frac{1}{x^2}-x^2}+e^{-\frac{1}{x^2}-x^2} \left(\frac{2}{x^3}-2 x\right) x\)

Using Mathematica I find two horizontal tangent lines at

\(\displaystyle x=\pm \sqrt{\frac{1}{4}+\frac{\sqrt{17}}{4}}\)

and also a horizontal tangent at \(\displaystyle x = 0\). (Note that it is tangent even though it crosses the curve; I had to look up the definition to be sure. For example, the function \(\displaystyle g(x)=x^3\) has a horizontal tangent at \(\displaystyle x=0\).)
 

dwsmith

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Mar 2010
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Florida
\(\displaystyle \frac{d}{dx}\left [xe^{-x^2-x^{-2}}\right]=e^{-x^2-x^{-2}}+xe^{-x^2-x^{-2}}(-2x+2x^{-3})=e^{-x^2-x^{-2}}(1+2x^{-2}-2x^2)\)

I added instead of subtracted the exponent when taking the derivative of e.

\(\displaystyle 2x^2+\frac{2}{x^2}+1=\frac{2x^4+x^2+2}{x^2}\)

\(\displaystyle x^2=y\)

\(\displaystyle 2y^2+y+2=\frac{-1\pm\sqrt{1-4(2)(2)}}{4}\)

Hmmm... What went wrong here.

I am not using a calculator for this problem since it is part of GRE Math subject test practice and I won't be allowed one on the test.
 

undefined

MHF Hall of Honor
Mar 2010
2,340
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Chicago
\(\displaystyle \frac{d}{dx}\left [xe^{-x^2-x^{-2}}\right]=e^{-x^2-x^{-2}}+xe^{-x^2-x^{-2}}(-2x+2x^{-3})=e^{-x^2-x^{-2}}(1+2x^{-2}-2x^2)\)

I added instead of subtracted the exponent when taking the derivative of e.

\(\displaystyle 2x^2+\frac{2}{x^2}+1=\frac{2x^4+x^2+2}{x^2}\)

\(\displaystyle x^2=y\)

\(\displaystyle 2y^2+y+2=\frac{-1\pm\sqrt{1-4(2)(2)}}{4}\)

Hmmm... What went wrong here.

I am not using a calculator for this problem since it is part of GRE Math subject test practice and I won't be allowed one on the test.
Okay, so I'm forced not to be lazy. :D

You changed a negative sign to a positive sign when copying from one line to the next. You should have

\(\displaystyle \color{red}-\color{black}2x^2+\frac{2}{x^2}+1\)
 
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dwsmith

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I hate when that happens.