Derivatives 4

1) Applying Mean Value Thm, for all 0 < x < 3:

[f(x) - f(0)]/(x - 0) = f'(c) where 0 < c < x

Hence

0 <= [f(x) - f(0)]/x <= 2

0 <= f(x) - f(0) <= 2x

For f(3), we have:

[f(3) - f(0)]/3 = f'(m) where 0 < m < 3

Again 0 <= [f(3) - f(0)]/3 <= 2

0 <= f(3) - f(0) <= 6

So 0 <= f(x) - f(0) <= 2x for all 0 < x <= 3

2) Construct a func:

f(x) = (1 + x)α - 1 - αx

f(0) = 0

f('x) = α(1 + x)α-1 - α

= α[(1 + x)α-1 - 1]

> 0 for any x > 0 since (1 + x)α-1 > 1 for the reason α > 1

Hence for any x > 0, there exists 0 < c < x such that:

[f(x) - f(0)]/x = f'(c) > 0

f(x) - f(0) > 0

f(x) > 0

(1 + x)α - 1 - αx > 0

(1 + x)α > 1 + αx