Use the Mean Value Theorem to establish the following inequalities:

(you may assume any relevant derivative formulas from calculus)

1. e^x > 1 + x for x>0

2. (x - 1) / x < lnx < x - 1 for x > 1

1) Applying the MVT to f(t) = e^t on [0, x] yields
(f(x) - f(0))/(x - 0) = f '(c) for some c in (0, x).
==> (e^x - 1)/x = e^c for some c in (0, x).

However, since e^x is increasing for all x, we have e^c > e^0 = 1 for all c in (0, x).
So, (e^x - 1)/x > 1 for x > 0
==> e^x > 1 + x for x > 0.
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2) This is done similarly.

Applying the MVT to f(t) = ln t on [1, x] yields
(f(x) - f(0))/(x - 1) = f '(c) for some c in (1, x).
==> (ln x - 0)/(x - 1) = 1/c for some c in (1, x).

However, since 1/x is decreasing for all x > 0, we have 1/x < 1/c < 1/1 for all c in (1, x).
So, 1/x < (ln x - 0)/(x - 1) < 1 for x > 1
==> (x - 1)/x < ln x < x - 1 for x > 1.

I hope this helps!