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...
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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`
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).`