Hm, it seems like his answer is off. The image is what I got.
Simplify in parentheses 5 radical three squared
1 answer:
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Answer:
N(1)=50 is a minimum
N(15)=4391.7 is a maximum
Step-by-step explanation:
<u>Extrema values of functions </u>
If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.
We are given a function (corrected)
(a)
First, we take its derivative
Solve N'(t)=0
Simplifying
Solving for t
Only t=1 belongs to the valid interval
Taking the second derivative
Which is always positive, so t=1 is a minimum
(b)
N(1)=50 is a minimum
(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15
(d)
N(15)=4391.7 is a maximum
Answer:
C) 6 inches
Step-by-step explanation:
Given function:
The function compares plant height in inches to the age of plant in weeks.
So, we can say
represents height of plant in inches.
represents age of the plant in weeks.
In order to predict the height of the plant in 5 weeks, we will plugin in the given function and evaluate
⇒
⇒
⇒ inches
Thus the approximate height of plant in 5 weeks would be inches