SOLUTION: Explain why there does not exist a polynomial p such that p(x) = 2^x for every real number x. [Hint: Consider the behavior of p(x) and 2^x for x near−∞.]

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Explain why there does not exist a polynomial p such that p(x) = 2^x for every real number x. [Hint: Consider the behavior of p(x) and 2^x for x near−∞.]       Log On


   

Question 950196: Explain why there does not exist a polynomial p such that p(x) = 2^x for every real number x. [Hint: Consider the behavior of p(x) and 2^x for x near−∞.]
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!



y+=+a%5Ex where a+%3E+0, the graph will asymptotically approach the line y+=+0 as x goes to -infinity, and y goes to infinity as x goes to infinity
you have p%28x%29+=+2%5Ex -> a=2%3E0, so
lim%28x-%3E-infinity%29 then 2%5Ex+=+0