1. ## [SOLVED] Solution Set

Find the solution set of $x + \frac{1}{4x} < 1$. The answer they gave was {x : x < 0}

2. Assume $x>0$
Then by AM-GM inequality,
$x+\frac{1}{4x}\ge 2*\sqrt{x*\frac{1}{4x}}$
$x+\frac{1}{4x}\ge 1$

So, $x<0$.

3. Originally Posted by mark1950
Find the solution set of $x + \frac{1}{4x} < 1$. The answer they gave was {x : x < 0}
$x+\frac{1}{4x}-1<0$

$\frac{4x^2}{4x}+\frac{1}{4x}-\frac{4x}{4x}<0$

$\frac{4x^2-4x+1}{4x}<0$

$
\frac{(2x-1)^2}{4x}<0
$

Note that this $(2x-1)^2$ is always positive so now lets look at the denominator $(4x)$ so the only possible solution here is $x<0$

4. Wow. thanks. How about if the numerator was $(2x - 1)^3$ as in always negative? What would the new solution set be? Thanks.

5. hi
$\left(2x-1 \right)^{3} = \left(2x-1 \right)^{2} \times \left(2x-1 \right)$
study the sign of $\frac{2x-1}{4x}$

6. Originally Posted by mark1950
Wow. thanks. How about if the numerator was $(2x - 1)^3$ as in always negative? What would the new solution set be? Thanks.

If its $\frac{(2x-1)^3}{4x}<0$

$\frac{(2x-1)^2(2x-1)}{4x}<0$

So you need to consider only $\frac{2x-1}{4x}<0$

so the solution will be (0,1/2)