Inequalities for f(x) is always positive Given that $f(x)=4x^2-1$
Find the range of values of $x$ so that $f(x)$ is always positive. 
My attempt,
$4x^2-1>0$
$4x^2>1$
$x^2>\frac{1}{4}$
$x>\pm\frac{1}{2}$
So $x<-\frac{1}{2}$ or $x>\frac{1}{2}$
Am I correct ?
 A: Yes indeed you are correct. 
Here is another solution just for confirmation: 
$4x^2 -1 = (2x-1)(2x+1) > 0$
(1) This occurs when both are positive:
$ 2x -1 >0$ when $x > 1/2$
$ 2x +1 >0$ when $x > -1/2$
Therefore, the solution here is $x > 1/2$
(2) This also occurs when both are negative:
$ 2x -1 <0$ when $x < 1/2$
$ 2x +1 <0$ when $x < -1/2$
Therefore, the solution here is $x < -1/2$
Giving your solution set of $ x< -1/2$ 
or $ x > 1/2$
Your method, however, is indeed the preferred one. 
Another option to consider is to find the roots quickly as $1/2$ and $-1/2$, realize the function opens upwards, and quickly come to the same conclusion as previously defined. 
A: Yes, you are correct. In cases such as these, where it is very easy to sketch the (quadratic) function, you are encouraged to do so, marking the roots. 

From there, you can easily see that your quadratic is positive after your positive root, and negative before your negative root. So $x > \frac{1}{2}$ and $x < -\frac{1}{2}$ is the desired solution. 

Note: the quadratic is u-shaped because the coefficient of the $x^2$ term is positive. Can you solve the same variant of this problem with the sketch for $f(x) = -4x^2 + 1$? Hint: It's not u-shaped anymore. 
