Question

When a new cellphone is put on the market, the demand each month can be described by the function C of t is equal to negative square root of the quantity t squared plus 4 times t minus 12 end quantity plus 3 where C (t) represents the demand of the cellphone (measured in millions of people) and the time, t, is measured in months. Which of the following solution(s) are valid for a positive demand?

Answer 1

A function is positive where it is above the x-axis

The valid solution for positive demand are; t = 3, and t = 2

The reason the above values are correct is as follows:

Known parameters:

The given function of the demand is; $C(t) = \mathbf{ -\sqrt{t^2 + 4 \times t - 12} +3}$

Where;

C(t) = The demand of the cellphone (in millions of people)

t = The number of months

The condition positive demand is C(t) ≥ 0

Therefore;

$-\sqrt{t^2 + 4 \times t - 12} +3 \geq 0$

$-\sqrt{t^2 + 4 \times t - 12} \geq -3$

$\sqrt{t^2 + 4 \times t - 12} \leq 3$

t² + 4·t - 12 ≤ 9

t² + 4·t - 12 - 9 ≤ 0

t² + 4·t - 21 ≤ 0

(t - 3) × (t + 7) ≤ 0

∴ t ≤ 3, or t ≥ -7

At t = 2 < 3, we have;

C(2) = -√(2² + 4×2 - 12) + 3 = 3

At t = 1 < 3, the function is; C(1) = -√(1² + 4×1 - 12) + 3 (Is undefined)

Therefore, the valid solution for positive demand are;

t = 3, and t = 2

Learn more about the functions here:

https://brainly.com/question/24249596

Answer 2

Answer:

3,3

Step-by-step explanation: