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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 minus 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?



(2, 3)

(6, 3)

(3, 0) and (7, 0)

(–2, 3) and (6, 3)

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 squar class=

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Answer:

Step-by-step explanation:

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;

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;

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

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