Curly question (1 Viewer)

no_arg

Member
Joined
Mar 25, 2004
Messages
67
Gender
Undisclosed
HSC
N/A
Prove that y=x^3 is an increasing function over the entire real line.
 

Nodice

Member
Joined
Feb 20, 2005
Messages
85
Location
Sydney
Gender
Male
HSC
2006
Hmm, isnt that just dy/dx = 3x^2 which is always >0, therefore an increasing function?
 

insert-username

Wandering the Lacuna
Joined
Jun 6, 2005
Messages
1,226
Location
NSW
Gender
Male
HSC
2006
The derivative may be 0 at the point (0,0), but the function is still increasing, since the definition of an increasing function is that f(b) > f(a) for all b>a. In this instance, the f(0) is larger than the f(-0.00000000000001, etc.), and the f(0.0000000,1 etc) is greater than the f(0), so it's an increasing function, even if dy/dx is 0 at one point.


I_F
 

acmilan

I'll stab ya
Joined
May 24, 2004
Messages
3,989
Location
Jumanji
Gender
Male
HSC
N/A
Lazarus said:
What about at x = 0?
the derivative method works for all x but 0, so i guess you'd take 0 as a special case and do something like:

Let a > 0 be any real number as small or large as you want and let f(x) = x3

f(0) = 0
f(0 + a) = a3 > 0 = f(0)
f(0 - a) = -a3 < 0 = f(0)

So f(-a) < f(0) < f(a)

which preserves the increasing function property.
 

no_arg

Member
Joined
Mar 25, 2004
Messages
67
Gender
Undisclosed
HSC
N/A
You need to show that
a less than b implies a^3 less than b^3
Best not to use calculus
 
Last edited:

Raginsheep

Active Member
Joined
Jun 14, 2004
Messages
1,227
Gender
Male
HSC
2005
for a<b,
as a approaches negative infinity, a^3 approaches negative infinity
as b approaches positive infinity, b^3 approaches positive infinity
since y=x^3 is an injective function,
then y=x^3 is increasing for all real x.
 

SeDaTeD

Member
Joined
Mar 31, 2004
Messages
571
Gender
Male
HSC
2004
Suppose b > a

b - a > 0
b^2 + ab + a^2 >= b^2 - |ab| + a^2
= (|b| - |a|)^2 +|ab| > 0 (equality only occurs with a=b=0, so this is strict)
therefore, (b-a)(b^2 + ab + a^2) >= 0
b^3 - a^3 >= 0
b^3 >= a^3.
 

Templar

P vs NP
Joined
Aug 11, 2004
Messages
1,979
Gender
Male
HSC
2004
SeDaTeD said:
Suppose b > a

b - a > 0
b^2 + ab + a^2 >= b^2 - |ab| + a^2
= (|b| - |a|)^2 +|ab| > 0 (equality only occurs with a=b=0, so this is strict)
therefore, (b-a)(b^2 + ab + a^2) >= 0
b^3 - a^3 >= 0
b^3 >= a^3.
Good...now use Daners' favourite method of proving from axioms. :p
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top