1. ## Decreasing function

$f(x)=x^\frac{1}{n} - (x-1)^\frac{1}{n}$

I need help showing this is decreasing for $x$> $1$

Any suggestions? trying to do this with the derivative.

I get

$f'(x)=\frac{x^\frac{1-n}{n}-(x-1)^\frac{1-n}{n}}{n}$

Can I assume $x^\frac{1-n}{n}<(x-1)^\frac{1-n}{n}$, If so why?

Any help would greatly be appreciated.

2. Originally Posted by hockey777
$f(x)=x^\frac{1}{n} - (x-1)^\frac{1}{n}$

I need help showing this is decreasing for $x$> $1$

Any suggestions? trying to do this with the derivative.

I get

$f'(x)=\frac{x^\frac{1-n}{n}-(x-1)^\frac{1-n}{n}}{n}$

Can I assume $x^\frac{1-n}{n}<(x-1)^\frac{1-n}{n}$, If so why?

Any help would greatly be appreciated.
Ok so we need to show that $x^{\frac{1}{n}}-(x-1)^{\frac{1}{n}}\geq(x+1)^{\frac{1}{n}}-(x)^{\frac{1}{n}}\forall{x}\in[1,\infty)$

Now we may rewrite this as

$x^{\frac{1}{n}}\geq\frac{(x-1)^{\frac{1}{n}}+(x-1)^{\frac{1}{n}}}{2}$

Or

$\left(\frac{(x+1)+(x-1)}{2}\right)^{\frac{1}{n}}\geq\frac{(x+1)^{\frac{ 1}{n}}+(x-1)^{\frac{1}{n}}}{2}$

and since

$\frac{d^2}{dx^2}\bigg[x^{\frac{1}{n}}\bigg]=\frac{x^{\frac{1}{n}}}{x^2}\bigg[\frac{1}{n^2}-\frac{1}{n}\bigg]$

Which is less than zero for all x and n greater than one...therefore $x^{\frac{1}{n}}$ is concave $\forall(x,n)\in\mathbb{R^+}$

Therefore $f\left(\frac{x+y}{2}\right)\geq\frac{f(x)+f(y)}{2} \quad\blacksquare$

3. I should be more clear that I'm trying to prove that

$a^\frac{1}{n}-b^\frac{1}{n}<(a-b)^\frac{1}{n}$

For $a>b>0$ and $n$>2[/tex]

by showing that function is decreasing, I can prove this. Therefore I can't use a few of your assumptions in example.

4. Originally Posted by hockey777
I should be more clear that I'm trying to prove that

$a^\frac{1}{n}-b^\frac{1}{n}<(a-b)^\frac{1}{n}$

For $a>b>0$ and $n$>2[/tex]

by showing that function is decreasing, I can prove this. Therefore I can't use a few of your assumptions in example.
I did show that your function is decreasing

5. Isn't what you did show that

f(x+1)<f(x)

not that $f(x_2)$< $f(x_1)$