The dollar value of a stock x weeks after a change in management is give by the function: f(x) = x²-14x+30ln(x+1)+32, 0≤x≤7. Find the intervals when the value of the sock was increasing and also determine the maximum value of the stock during this time. PLEEASEE HELP ME!!! and please show the work so that I can understand how to do this!! It truly means the world to me, thank you so much!!!
To determine the intervals when the value of the stock is increasing, we need to analyze the derivative of the function f(x) with respect to x. If the derivative is positive, it means the function is increasing in that interval.
Let's begin by finding the derivative of f(x):
f'(x) = d/dx(x²) - d/dx(14x) + d/dx(30ln(x+1)) + d/dx(32)
f'(x) = 2x - 14 + (30/(x+1))(1) + 0
Simplifying further:
f'(x) = 2x - 14 + 30/(x+1)
To determine when the stock value is increasing, we need to find the values of x that make f'(x) positive.
2x - 14 + 30/(x+1) > 0
To solve this inequality, we can break it down into two cases:
Case 1: When x+1 > 0 (i.e., x > -1)
2x - 14 + 30/(x+1) > 0
Multiplying through by (x+1) to eliminate the denominator:
(2x - 14)(x+1) + 30 > 0
Simplifying further:
2x² - 12x + 16 > 0
We can use the quadratic formula to find the solutions:
x = (-(-12) ± √((-12)² - 4(2)(16))) / (2(2))
x = (12 ± √(144-128)) / 4
x = (12 ± √16) / 4
The solutions are:
x = (12 + 4) / 4 = 4
x = (12 - 4) / 4 = 2/4 = 1/2
Since x + 1 > 0, we keep the solution x = 4 and exclude x = 1/2.
Case 2: When x+1 < 0 (i.e., x < -1)
2x - 14 + 30/(x+1) > 0
Multiplying through by (x+1) and reversing the inequality:
(2x - 14)(x+1) + 30 < 0
Simplifying further:
2x² - 12x + 16 < 0
Using the quadratic formula again:
x = (12 ± √(144-128)) / 4
x = (12 ± √16) / 4
The solutions are:
x = (12 + 4) / 4 = 4
x = (12 - 4) / 4 = 2/4 = 1/2
Since x + 1 < 0, we exclude these solutions.
Now let's determine the maximum value of the stock during this time. To find the maximum, we need to evaluate the function f(x) at the critical points and endpoints of the interval [0, 7].
f(0) = (0)² - 14(0) + 30ln(0+1) + 32 = 32
f(7) = (7)² - 14(7) + 30ln(7+1) + 32 = 49 - 98 + 30ln(8) + 32 ≈ 72.335
f(4) = (4)² - 14(4) + 30ln(4+1) + 32 ≈ 59.204
f(1/2) = (1/2)² - 14(1/2) + 30ln(1/2+1) + 32 ≈ 65.508
The maximum value of the stock occurs at x = 4, and its value is approximately 59.204.
Therefore, the stock value is increasing in the interval [0, 4) and the maximum value of the stock during this time is approximately 59.204.