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 = ...?

Expert Answers

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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`


`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).`

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