Question

How do I use a sign chart to solve `2< -16t^2+6<5`?

Answer 1

`2 < -16t^2+6 < 5` in the intervals (-0.5, -0.25) and (0.25, 0.5).

Explanation:

Here, we have two inequalities that must be satisfied.

(1) `2 < -16t^2+6` and
(2) `-16t^2+6 < 5`

(1) The first inequality

We write the inequality in standard form by putting all non-zero terms on the left side.

`16t^2-6 +2 < 0`
`16t^2-4 < 0`

We start by finding the critical numbers.

Set `f_1(t) = 16t^2-4 = 0` and solve for `t`.

`4t^2-1 = 0`

`(2t+1)(2t-1) = 0`

`2t+1 = 0` or `2t-1 = 0`

`t=-0.5` or `t=+0.5`

The critical numbers are `-0.5` and `+0.5`.

We have three intervals to consider: (-∞, -0.5), (-0.5, 0.5), and (0.5, ∞).

We pick a test number and evaluate the function and its sign at that number.

(2) The second inequality

The inequality in standard form is

`-16t^2+6 -5 < 0`
`-16t^2+1 < 0` or `1-16t^2 <0`

Set `f_2(t) = 1-16t^2 = 0` and solve for `t`.

`(1+4t)(1-4t) = 0`

`1+4t = 0` or `1-4t = 0`

`t=-0.25` or `t=+0.25`

The critical numbers are `-0.25` and `+0.25`.

The three intervals to consider are: (-∞, -0.25), (-0.25, 0.25), and (0.25, ∞).

Now we create a sign chart for the two functions.

The only intervals for which the two signs are both negative are (-0.5, -0.25) and (0.25, 0.5).