1. ## [SOLVED] Solution Set

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

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

So, $\displaystyle x<0$.

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

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

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

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

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

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

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

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

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

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

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

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