# please help me to solve this: f ' (x) < 0 for x > 0, f (2 x squared + x + 1) < f (3 x squared - 4x +1 ), X = ...?

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### 2 Answers

The problem provides the information that `f'(x)lt0` , hence the function decreases over the interval `(0,oo).`

You need to remember that a decreasing function follows the rule:

`x_1ltx_2 =gt f(x_1)gtf(x_2)`

The problem provides the information that `f(2x^2+x+1)ltf(3x^2-4x+1)` , hence `2x^2+x+1` needs to be larger than `3x^2-4x+ 1` thus you need to solve the inequality `2x^2+x+1 gt 3x^2-4x+1` to find x such that:

`2x^2+x+1 gt 3x^2-4x+1`

`2x^2 + x + 1- 3x^2 + 4x - 1 gt 0`

Reducing like terms yields:

`-x^2 + 5x gt 0`

You need to multiply by -1 such that:

`x^2 - 5x lt 0`

You need to factor out x such that:

`x(x - 5)lt0`

Hence, the factors x and x - 5 need to have different signs for the product to be negative such that:

`x gt 0`

`x - 5 lt 0 =gt x lt 5`

or

`x lt 0`

`x - 5 gt 0 =gt x gt 5`

**Since the intersection of the intervals `(-oo,0)` and `(5,oo)` yields `O/` , hence the common values for x that satisfy the inequality `x(x - 5)lt0` are in interval `(0,5).` **

I'm not sure I understand the whole question, but here's something to think about:

If f'<0 the function is decreasing. Another way to put that is:

if A<B, f(A)>f(B)

Your two arguments for f are 2x^2+x+1 and 3x^2-4x+1. These two intersect at x=0 and x=5. Between 0 and 5:

2x^2+x+1 >3x^2-4x+1

So

f(2x^2+x+1) < f(3x^2-4x+1)

So

0<x<5

What bothers me about this is that restriction on f', that we only know f'<0 when x>0. Or are you saying you only want solutions where x>0?