we will select each options and find zeros
(a)
for finding x-intercept , we can set g(x)=0
and then we can solve for x
now, we can factor it
we get
so, this is TRUE
(b)
we can set it to 0
and then we can solve for x
we get
so, this is FALSE
(c)
we can set it to 0
and then we can solve for x
so, this is TRUE
(d)
now, we can set it to 0
and then we can solve for x
so, this is TRUE
(e)
we have
now, we can set it to 0
and then we can solve for x
so, this is FALSE
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