A brand new stock is also called an initial public offering. In this model, the period immediately after the stock is issued offers excess returns on the stock—that is, the stock is selling for more than it is really worth. One such model predicts the percent overvaluation of a stock as R(t)=9t((t-3)^3/2.718)
where R is the overvaluation in percent and t is the time in months after the initial issue.
Use the information provided by the first derivative, second derivative to prepare advice for clients as to when they should expect a signal to prepare to buy or sell (inflection point), the exact time when they should buy or sell (local maximum/minimum), and any false signals. Explain your reasoning.
Would the first derivative be R'(t)=3.311258278(3(t-4)^2(t)+(t-4)^3) and the second derivative be R''(t)=3.311258278(6(t-4)x+6(t-4)^2)?
The way you typed R(t), we have to read it as
R(t) = 9/2.718 ( t(t-3)^3)
then R' (t) = (9/2.718) [ t (3)(t-3)^2 + (t-3)^3 ]
= (500/151)(t-3)^2 ( 3t + t-3)
= (500/151)(4t - 3)(t-3)^2
How did the base of (t-3) suddenly turn into (t-4) ?
Sorry about that. It looks like thins:
R(t)= 9t[(t-4)^3/2.718]
The 9t is on the outside without a denominator
Yes, you are correct in calculating the first derivative and second derivative of the function R(t).
To confirm, let's go through the calculations step by step:
1. First derivative (R'(t)):
- Take the derivative of each term in the function R(t) using the power rule and chain rule.
- We obtain R'(t) = 3.311258278 * (3(t-4)^2(t) + (t-4)^3)
2. Second derivative (R''(t)):
- Take the derivative of R'(t) obtained in the previous step.
- We obtain R''(t) = 3.311258278 * (6(t-4)x + 6(t-4)^2)
Now, let's use this information to provide advice to clients:
1. Inflection Point:
- An inflection point is where the second derivative changes its sign, indicating a change in the concavity of the function.
- To find the inflection point, we need to solve the equation R''(t) = 0 for t.
2. Local Maximum/Minimum:
- A local maximum or minimum occurs where the first derivative changes its sign, indicating either an increase or decrease in the function.
- To find the local maximum/minimum, we need to solve the equation R'(t) = 0 for t.
3. False Signals:
- False signals can occur when the first derivative is zero (local maximum/minimum) but the second derivative is not.
- To check for false signals, evaluate the sign of the second derivative at the local maximum/minimum points. If it is the same on both sides, it indicates a potential false signal.
By analyzing the first derivative and second derivative, you can determine the inflection points, local maximum/minimum points, and identify any potential false signals.