D

Let a be a real number, 0≤a≤1. Show that (1+x)^a ≤ 1+ ax for x -1.

11 viewed last edited 1 year ago
Jada Hunt
0
This is probably a lot easier than i think it is. This probably solves itself by showing the relationship between the derivative of LHS and RHS, but I was a little stumped by the multiple variables. Or do i show that there does not exists a number 'a' in the given interval, where (1+x)^a > 1+ax?
Mahesh Godavarti
1
I will show you how to prove this using derivatives. You can take the derivative with respect to either variable treating the other as a constant. Let's first treat a as a constant and take the derivate with respect to x . Let the expression on the left hand side (LHS) be f(x) and the expression on the right hand side (RHS) be g(x) . Then f'(x) = a (1+x)^{a-1} and g'(x) = a . Now take the second derivate with respect to x which gives f''(x) = a (a-1) (1+x)^{a-2} and g''(x) = 0 . Now, note that f''(x) < 0 = g''(x) as for 0 < a < 1 and x > -1 , we have a > 0, (1+x) > 0 \text{ and } (a-1) < 0 . Since f'(-1) = 0 < a = g'(-1) and f''(x) < g''(x) for all x > -1 , we have f'(x) < g'(x) for all x > -1 . Since f(-1) = 0 < 1 - a = g(-1) and f'(x) < g'(x) for all x > -1 , we have f(x) < g(x) for all x > -1 . Q.E.D.