Analysis Problems #5
Solutions
1.
Show that the polynomial f ( x ) = x 3 −
4 x 2 −
3 x + 1 has exactly one root in [0 , 2] .
•
Being a polynomial, f is continuous on the closed interval [0 , 2] and we also have f (0) = 1 > 0 , f (2) =
−
13 < 0 .
Thus, f has a root in (0 , 2) by Bolzano’s theorem and this root certainly lies in [0 , 2] as well. Suppose now that f has two roots in [0 , 2]. By Rolle’s theorem, f have a root in [0 , 2]. However, the roots of f
′
( x ) = 3 x 2 −
8 x
−
3 are x =
8
±
√
64 + 4
·
3
·
3
2
·
3
=
8
±
10
6
=
⇒ x = 3 , x =
−
1 / 3 .
′ must then
Since neither of those lies in [0 , 2], we conclude that f cannot have two roots in [0 , 2].
2.
Suppose log x is a function such that (log x )
′
=
1 x and log 1 = 0 . Show that log x
≤ x
−
1 for all x > 0 .
•
We compute the maximum value of f ( x ) = log x
− x + 1. Since the derivative f
′
( x ) =
1 x
−
1 =
1
− x x is positive when 0 < x < 1 and negative when x > 1, this function is increasing for the former values of x and decreasing for the latter. In particular, f (1) = log 1
−
1 + 1 = 0 is the maximum value attained by the function, so f ( x )
≤ f (1) = 0 for all x > 0.
3.
Suppose that x > y > 0 . Use the mean value theorem to show that
1
− y x
< log x
− log y < x y
−
1 .
•
Since f ( x ) = log x is differentiable with f
′
( x ) = 1 /x , the mean value theorem gives f
′
( c ) = f ( x )
− f ( y ) x
− y
=
⇒
1 c
= log x
− log y x
− y for some x > c > y . Using this fact, we now get x > c > y =
⇒
1 x
<
1 c
<
1 y
=
⇒
1 x
< log x
− log y x
− y
1
< y so we can multiply through by x
− y > 0 to deduce the desired inequality.

4.
Suppose that f is twice differentiable with f (0) = f
′
(0) = 0 and f
′′
( x ) + f ( x ) = 0 for all x
∈ R
.
Show that f ( x )
2
+ f
′
( x )
2
= 0 for all x
∈ R
. Deduce that f ( x ) = 0 for all x
∈ R
.
•
Setting g ( x ) = f ( x ) 2 + f
′
( x ) 2 for convenience, we get g
′
( x ) = 2 f ( x ) f
′
( x ) + 2 f
′
( x ) f
′′
( x ) = 2 f
′
( x )
·
[ f ( x ) + f
′′
( x )] = 0 .
This shows that g ( x ) is actually constant, so f ( x )
2
+ f
′
( x )
2
= g ( x ) = g (0) = f (0)
2
+ f
′
(0)
2
= 0 for all x
∈ R
. Since the sum of two non-negative terms can only be zero when both terms are zero, we deduce that f ( x ) = 0 for all x
∈ R
.
5.
Let n
∈ N be fixed. Show that x n log x
≥ − 1 ne for all x > 0 .
•
Setting f ( x ) = x n log x for convenience, we can use the product rule to get f
′
( x ) = nx n
−
1 · log x + x n · x
−
1
= x n
−
1
( n log x + 1) .
Since x > 0 by assumption, this expression is negative when n log x + 1 < 0
⇐⇒ log x <
−
1 /n
⇐⇒ x < e
−
1 /n and positive when x > e
−
1 /n . In particular, f is decreasing for the former values of x and increasing for the latter, so f attains its minimum value when x = e
−
1 /n and f ( x )
≥ f ( e
−
1 /n
) =
( e
−
1 /n
) n · log e
−
1 /n
= e
−
1 ·
(
− n
1
)
=
−
1 ne
.