Question

The price of a certain computer stock t days after it is issued for sale is p ( t ) = 100 + 20t − 3t^2 dollars. The price of the stock initially rises, but eventually begins to fall. During what period of time does the stock price rise? 0 ≤ t ≤ If you owned the stock, after how many days would you sell it? days

Answer 1

The period of time does the stock price rise is 0 ≤ t ≤ 10/3.

It would sell in 3.34 days.

What is differentiation?

It is the process of finding the derivative, or rate of change, of a function.

The price of a certain computer stock after t days is

p(t) = 100 + 20t - 3t²

Now, take the derivative of the given function and equate it to zero to find the critical points,

p'(t) = 20 - 6t = 0

t = 20/6

t = 10/3 days

Thus, there are two intervals in which the given function is defined

(0, 10/3) and (10/3, ∞)

For the interval (0,10/3 ),

p'(1) = 20 - 6(1) =14

For the interval (10/3, ∞),

p'(2) = 20 - 6(2) = 8

Since p'(t) in the interval (0, 10/3) indicates that the function is increasing.

0 < t <10/3

Since at the point t = 3.34 days it would sell.

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