The base case doesn't have to be n = 1. Use n = 6 for your base case.mebigp said:Homework Statement
Given as true f(x) = (1+1/x)^x is strictly increasing for x>=1 and that f(x) has horizontal asymptote y=e.
Prove that (n/3)^n< n! <(n/2)^n for all integers n>=6 ?
Homework Equations
The Attempt at a Solution
f(x)=(1+1/x)^x is increasing and approach e
prove (n/3)^n< n! <(n/2)^n for all n>=6
So I attempt to use f(x) to replace n but that will not work for the base case 1 because n>6
A function is strictly increasing if the values of the function increase as the input values increase. In other words, as the x-values increase, the corresponding y-values also increase.
We can prove that f(x) = (1+1/x)^x is strictly increasing for x>=1 by taking the derivative of the function and showing that it is always positive for x>=1. This means that the function is always increasing and never decreasing for all values of x greater than or equal to 1.
The restriction x>=1 is necessary because the function (1+1/x)^x is only strictly increasing for values of x greater than or equal to 1. For values of x less than 1, the function is not strictly increasing and can even be decreasing at certain points.
Yes, the graph of f(x) = (1+1/x)^x will show a steadily increasing curve for all values of x greater than or equal to 1. This means that as x increases, the corresponding y-values on the graph will also increase, demonstrating the strict increasing nature of the function.
Yes, there are other ways to prove this statement. One way is to use the mean value theorem, which states that if a function has a positive derivative for all values in an interval, then the function must be strictly increasing on that interval. Another way is to use the definition of a strictly increasing function and show that it holds true for f(x) = (1+1/x)^x for all x-values greater than or equal to 1.