Increase/Decrease of Functions Problem

[turophile]

Here's the problem:

Show that if s is any rational number, s > 1, then (1 + x)[sup:3tqqqfnp]s[/sup:3tqqqfnp] > 1 + sx for all x > 0.

The textbook gives the hint that, if c is any number > 1 and t is any rational number > 0, then c[sup:3tqqqfnp]t[/sup:3tqqqfnp] > 1. I'm not quite sure how to get started with this one. Any other hints?

 

[Deleted member 4993]

Here's the problem:

Show that if s is any rational number, s > 1, then (1 + x)[sup:520un1zz]s[/sup:520un1zz] > 1 + sx for all x > 0.

The textbook gives the hint that, if c is any number > 1 and t is any rational number > 0, then c[sup:520un1zz]t[/sup:520un1zz] > 1. I'm not quite sure how to get started with this one. Any other hints?

Try Taylor series expansion

 

[turophile]

Taylor series expansion comes much later in the textbook, so I don't think I'm supposed to use it for this problem. It comes from a section that introduces the sign of the derivative, local maxima and minima, concavity and inflection, and second-derivative tests. Using other examples in the textbook, I think I need to set up the problem something like this:

Let g and h be continuous on an interval I and differentiable on its interior, and let a be a point of I. Let f = g - h. Then if g' - h' > 0 throughout the interior and if g(a) ? h(a), then g(x) > h(x) for all a in I.

Working backward, I'm trying to get to this last step in the proof:

(1 + x)[sup:3erjyrf3]s[/sup:3erjyrf3] - 1 - sx > 0 ? (1 + x)[sup:3erjyrf3]s[/sup:3erjyrf3] > 1 + sx

I'm just not sure how to set up g and h to get to this result.

 

[Deleted member 4993]

Okay then how about this:

x > 0

1+ x > 1

(1+x)[sup:2z73umbw]s[/sup:2z73umbw] > 1[sup:2z73umbw]s[/sup:2z73umbw]

(1+x)[sup:2z73umbw]s[/sup:2z73umbw] > 1