# Given function f(x)=x/x^4+3, is true that absolute val(f(x)-f(y)) < absolute value (x-y), if all values x,y reals

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### 1 Answer

You need to prove the following inequality such that:

`|f(x) - f(y)|<|x - y|`

You may divide by `|x - y|` both sides such that:

`|f(x) - f(y)|/|x - y| < 1`

Notice that `|x - y| > 0` , hence, the inequality holds.

You need to remember the mean value theorem such that:

`|f(x) - f(y)|/|x - y| = f'(c)< 1` , where `c in (x,y)`

You need to prove that the maximum of the function is less than 1, hence, you need to find the derivative of th given function, using the quotient rule, such that:

`f'(x) = (x'(x^4+3) - x(x^4+3)')/((x^4+3)^2)`

`f'(x)= (x^4 + 3 - x(4x^3))/((x^4+3)^2)`

`f'(x) = (x^4 + 3 - 4x^4)/((x^4+3)^2)`

`f'(x) = (-3x^4 + 3)/((x^4+3)^2)`

You need to factor out `3` such that:

`f'(x) = 3(1 - x^4)/((x^4+3)^2)`

You need to solve for `x` the equation `f'(x) = 0` to evaluate the critical points of the function f`(x)` such that:

`3(1 - x^4)/((x^4+3)^2) = 0`

`Since ((x^4+3)^2) != 0 => 3(1 - x^4) = 0`

Dividing by `3` yields:

`1 - x^4 = 0`

Converting the difference of squares into a product yields:

`(1 - x^2)(1 + x^2) = 0`

Since the equation `1 + x^2 = 0` has no real roots, hence `1 - x^2 = 0 => x_(1,2) = +-1`

You need to evaluate the maximum of the function at `x = 1` such that:

`f(1) = 1/(1^4 + 3) = 1/4 < 1`

**Hence, checking the inequality `|f(x) - f(y)|<|x - y|` yields that it holds for reals x and y.**