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 = ...?
2 Answers
sciencesolve | Teacher | (Level 3) Educator Emeritus
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).`
baronridiculous | College Teacher | (Level 1) eNoter
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?