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Intersection of two functions 
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#1
Oct1708, 09:57 PM

P: 64

(Problem from practice math subject GRE exam:) At how many points in the xyplane do the graphs of [tex]y=x^{12}[/tex] and [tex]y=2^x[/tex] intersect?
The answer I got was 2, but the answer key says 3. Intuitively, by the shape of their graphs, I would say two. I tried to calculate actual values for x: [tex]2^x=x^{12}[/tex] [tex]x\ln2=12\ln x[/tex] [tex]\frac{\ln2}{12}=\frac{\ln x}{x}[/tex] [tex]\sqrt[12]{2}=\sqrt[x]{x}[/tex] I don't know what to do with that last equation. I'm really confused though, because I can't even imagine how they would get a third intersection. Any help would be appreciated. :) 


#2
Oct1808, 02:31 AM

P: 336

It's pointless to try to solve a transcendental equation analytically. Remember that x^12 is an even function, and note that 2^x approaches 0 as x approaches infinity, but also remember that when x=0 that x^12 = 0, so you know that the two plots cross once for x < 0. You might guess they cross once for x>=0, but think about when x is > say 1000 and when x is say 2. Which function is larger in each case? Which is larger at x=0? Which is larger for x = 1000?



#3
Oct1808, 05:33 AM

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PF Gold
P: 39,498

Clearly the graph of y= 2^{x} crosses the graph of y= x^{12} for x somewhere between 1 and 0: 2^{1}= 1/2 and (1)^{12}= 1 so the graph of x^{12} is higher for x= 1 while, at x= 0, 2[sup]0[sup]= 1 and 0^{12}= 0 so the graph of 2^{x} is higher for x= 0.
Also 2^{12}= 4096 while 2^{2}= 4: the graph of x^{12} is higher again so the graph must intersect again between x= 0 and x= 2. The question, then, is whether the graphs intersect a third time for x> 2; whether 2^{x} is larger than x^{12} for "sufficiently large x". One way to answer that is to look at the limit of 2^{x}/x^{12} as x goes to infinity. Since that fraction itself becomes "infinity over infinity" we can apply L'Hopital's rule. Repeatedly differentiating, the numerator just stays 2^{x} (times a power of ln(2)) while the denominator has lower and lower powers eventually becoming a constant (after 12 differentiations, we get 12!) and then 0. What does that tell you about the limit? And what does that tell you about whether 2^{x} or x^{12} is larger for very large x? 


#4
Oct1808, 10:59 AM

P: 64

Intersection of two functions
[tex]\lim_{x\rightarrow\infty}\frac{2^x}{x^{12}}=\lim_{x\rightarrow\infty}\f rac{(\ln2)2^x}{12x^{11}}=...=\lim_{x\rightarrow\infty}\frac{(\ln2)^{12} 2^x}{12!}=\infty[/tex] Which means that for very large x, 2^x does eventually exceed x^12, which gives us the third intersection point. So, one last question  Is this a good general strategy for this type of problem (if I were to get a similar one on the actual exam): First sketch the graph and see what obvious/immediate intersection points I can find. Then use the limit idea for [tex]x\rightarrow\infty[/tex] and [tex]x\rightarrow\infty[/tex]. Will this ensure that I find all of my intersection points? Thanks so much! :) 


#5
Oct1808, 12:35 PM

Math
Emeritus
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PF Gold
P: 39,498

Not necessarily. For example, it there were 3 more intersections between x= 2 and infinity, the same changes in which is smaller and which is larger would be true. You might try looking at the derivative of f g. If that is always positive, then that can't happen.



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