Attached are the solutions. Please let me know if you have any questions or if there are any errors.
Start by finding the derivative of the function:
f(x) = 5x^-2
f'(x) = 5*-2x^-3 = -10/x^3
Evaluate the function and the derivative at the point x = 1
f(1) = 5/1^2 = 5
f'(1) = -10/1^3 = -10
f'(1) will be the slope of the tangent line.
Use point slope form to develop the equation for the tangent line
y – y1 = m(x – x1)
y – 5 = -10(x-1)
y -5 = -10x + 10
y = -10x + 15
Evaluating at negative infinity would give you a indeterminate form. Take the derivative of the
top and bottom by applying l'hopitals
lim(x--> - inf) 14x + 10 / 12x – 5
Evaluating this limit at negative infinity would still give you in indeterminate form of – infinity / infinity. That means you should apply l'hopitals rule one more time
lim(x--> -inf) 14 / 12 = 14/12 = 7/6
The limit is 7/6
P'(4) was equal to 35. If t = 4, that means the year would be 2000 + 4 = 2004.
The rate would be 35000. Remember P is in thousands of dollars, so 35 really means 35000
Box 1: 35000
Box 2: 2004
Start with s(t) and set that equal to zero. That will tell you the time when the rock reaches the
0 = 400 – 16t^2
16t^2 = 400
t^2 = 25
Since you cannot use a negative time, you should focus on the t = 5
Now you also needs to find the velocity at that time by plugging t = 5 into your derivative
S'(5) = -32*5 = -160 ft / sec
Box 1: - 160
Box 2: 5
The revenue function can be fo...