# Mean value thereom application

1. Dec 10, 2016

### Arnoldjavs3

1. The problem statement, all variables and given/known data
Suppose that x,y are in (0,0.5). Show that:
$$\left|\frac{1}{x^3}-\frac{1}{y^3}\right| ≥ 48|x-y|$$

2. Relevant equations

3. The attempt at a solution
I think that we need a function $f(t) = t^{-3}$ & $f'(t) = -3t^{-4}$. Now we use the MVT here to show that there is some c that makes the expression in the problem correct right? My question is how do I incorporate the absolute value notations in the statement here?
$$-3c^{-4} = \frac{f(y) - f(x)}{y-x}$$
I think we have to assume that either x or y is greater than the other(because if they were equal then the above owuld be proven). Do I just input c for 0.5? I'm unsure.

2. Dec 10, 2016

### PeroK

It's not clear why the MVT would help here.

3. Dec 10, 2016

### Arnoldjavs3

I forgot to add that they are differentiable on the interval (x,y)

4. Dec 10, 2016

### PeroK

I still can't see why you think the MVT helps you.

5. Dec 10, 2016

### LCKurtz

Remember that all your variables are in $(0,\frac 1 2)$. If you take the absolute value of both sides of what you have written you get$$\frac 3 {c^4} =\frac{|f(y) - f(x)|}{|y-x |}$$which can be rewritten$$\frac 3 {c^4} |y-x| =\left | \frac 1 {x^3} - \frac 1 {y^3}\right |$$Think about how large or small the left side can be.

6. Dec 12, 2016

### Arnoldjavs3

Since c is an element of x,y then it can't be larger than 0.5 or smaller than 0? And as such this proves that since 48|y-x| will be greater since y is greater than c? So the side with c will always be smaller.

Last edited: Dec 12, 2016
7. Dec 12, 2016

### LCKurtz

That doesn't make any sense. $c$ is a number in $(0,.5)$, not an "element" of x,y, whatever that means.

Stating the result doesn't prove anything. You need to show some inequalities.

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