(Problem from practice math subject GRE exam:) At how many points in the xy-plane 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. :)
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?
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^{0= 1 and 012= 0 so the graph of 2x is higher for x= 0. Also 212= 4096 while 22= 4: the graph of x12 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 2x is larger than x12 for "sufficiently large x". One way to answer that is to look at the limit of 2x/x12 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 2x (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 2x or x12 is larger for very large x?}
Thanks! This makes sense. So... [tex]\lim_{x\rightarrow\infty}\frac{2^x}{x^{12}}=\lim_{x\rightarrow\infty}\frac{(\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! :)
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.