Curly question (1 Viewer)

[no_arg]

Prove that y=x^3 is an increasing function over the entire real line.

 

[Nodice]

Hmm, isnt that just dy/dx = 3x^2 which is always >0, therefore an increasing function?

 

[pLuvia]

Hmm, isnt that just dy/dx = 3x^2 which is always >0, therefore an increasing function?

Yep that should be right

 

[Lazarus]

What about at x = 0?

 

[insert-username]

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]

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]

You need to show that
a less than b implies a^3 less than b^3
Best not to use calculus

 

[Raginsheep]

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]

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]

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.