Calc. Practice Problems #11 of 16?

Question

Hello. I have just uploded onto Photobucket 16 calculus practice problems, all of which will be of similar content to my eventual final exam. Now, an answer to these questions would be nice, but is not necessary. Instead, I am concerned with and interested in THE METHOD as to how one could arrive to the correct answer, as well as an EXPLANATION as to why each step needs to be performed in relation to the ones that follow. Anybody who can break down and explain the contents of these problems to me in a SIMPLISTIC way would just be amazing in my book. So thank you very much should you decide to help enlighten me, as I appreciate this more then the words in this post could even begin to convey!

Question #11: http://www.i138.photobucket.com/albums/q271/Link3324/math11.jpg

By the way, This is NOT for any kind of required assignment. It is a guide to work from and to understand before the actual test. Any help would be greatly appreciated. THANK YOU!

Answer 1

first the above person who uses mathematical induction is correct however this a calculus class not introduction to number theory so i doubt you understood what he did.
here what i think you will uderstant
rewrite the equation as the hint states into
e^x-x-1>=0 take the taylor series expansion of e^x
expansion(e^x) = 1 + x + x^2/2! + x^3/3! +..... substitude this to
the above equation for e^x
1 + x + x^2/2! + x^3/3! +..... -x - 1 >= 0 reshuffle the order
1 -1 +x - x + x^2/2! + x^3/3! +..... >= 0 so 1 and x cancel
x^2/2! + x^3/3! +..... >= 0 clearly for x>=0 the equation is bigger than zero.

Answer 2

You want to use mathematical induction.
(/=/ means approximately equal to )
First for the basis step, you want to show that f(x)=e^x-1-x is greater than 0 for x=1.
f(1)=e^1-1-1 /=/ 2.718-2 /=/ 0.718.

Then for the induction step, you want to show that if given f(a)>=0, f(a+1)>=0 for all positive numbers a.
f(a+1) >=0 iff (if and only if) e^(a+1)-1-(a+1) >=0.
(e^a)*(e^1)-1-a-1>=0

Given e^a-1-a>=0 and since a>=1, e>=1>=0,
e*(e^a-1-a)>=e*0=0
e^(a+1)-1*e-a*e>=0
e^(a+1)>=1*e-a*e = e+ae >= 2+ae >= 2 + a (since ae>=a)
e^(a+1)>=2+a
e^(a+1)-2-a>=0
e^(a+1)-1-1-a>=0
e^(a+1)-1-(1+a)>=0
e^(a+1)-1-(a+1)>=0
Thus f(a) implies f(a+1).
This is only true for nonnegative integers, however.

A more complete solution can be seen by taking the suggested function f(x)=e^x-1-x and its derivative f'(x)=e^x-1. To find the extrema of f'(x), set the derivative equal to 0. f'(x)=e^x-1=0 or e^x=1 which gives x=0. (f(0)= e^0-1-0 = 0). Since e^x is an nonnegative function >=1 for all values of x on (0,infinity), so is e^x-1. Thus f'(x) is always increasing on (0,infinity), and your function f(x) is always increasing. Therefore, f(x)=e^x-1-x>=0 for all x on [0,infinity) and e^x>=1+x.