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

1 Answer
Jun 30, 2015

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.

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(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, ∞).

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Now we create a sign chart for the two functions.

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