T = 5, so after 5 years
p(t) = t^3 - 14t^2 + 20t + 120
Take derivative to find minimum:
p’(t) = 3t^2 - 28t + 10
Factor to solve for t:
p’(t) = (3t - 2)(t - 5)
0 = (3t - 2)(t - 5)
0 = 3t - 2
2 = 3t
2/3 = t
Plug 2/3 into original equation, this is a maximum. We want the minimum:
0 = t - 5
5 = t
Plug back into original:
5^3 - 14(5)^2 + 20(5) + 120
125 - 14(25) + 100 + 120
125 - 350 + 220
- 225 + 220
p(5) = -5
Answer:
I think its A
Step-by-step explanation:
Answer:
final answer is 6
Step-by-step explanation:
Given 
To find the difference quotient use formula
Now plug it in the formula
6
Answer:
Step-by-step explanation:
The third, fifth and sixth expressions (from the top) are NOT polynomials, and the reason in each case is that the expression has one or more negative powers of x in it.
