Answer:
k=16
Step-by-step explanation:
So the tangent line is

and it tangent to function

Since the slope of the tangent line is 4, this means the derivative of f(x) is 4 but first let find the derivative of

Use the Sum Rule,

Use the Power Rule and we get

Set this equal to 4



So at x=-2, the slope of the tangent line is 4.
Plug -2 in the orginal function, and we get

So the point must pass through -2,8 with a slope of 4.



So the value of k is 16.
Answer:
Part A: From 0 to 2 seconds, the height of the water balloon increases from 60 to 75 feet, therefore the water balloon's height is increasing during the interval [0,2]
Part B: From 2 to 4 seconds, the height of the water balloon stays the same at 75 feet, therefore the water balloon's height is the same during the interval [2,4] From 10 to 12 seconds, the height of the water balloon stays the same at 0 feet, therefore the water balloon's height is the same during the interval [10,12] From 12 to 14 seconds, the height of the water balloon stays the same at 0 feet, therefore the water balloon's height is the same during the interval [12,14]
Part C: The interval, [4,6] of the domain is when the water ballon's height decreases the fastest. The interval [4,6] decreases by 35 feet. The two other intervals that decrease are [6,8] and [8,10] which both have the same slope. They decrease by 20 feet. Therefore, this helps us conclude that the interval [4,6] decreases the fastest because 35 feet is a more significant decrease than 20 feet.
Part D: I predict that the height of the water balloon at 16 seconds is 0 feet. This is because at 10-14 seconds, the water balloon's height is 0 feet. In read-world situations, if the water balloon is on the ground which is 0 feet, it stays on the ground due to gravity.
Step-by-step explanation:
I hope this helps! I also do not know if it is all correct but I did research and everything so hopefully it is correct! Good luck!
Answer:
as x --> oo, f(x) = oo (infinite)
as x -- -oo, f(x) = 3
You're looking for the extreme values of
subject to the constraint
.
The target function has partial derivatives (set equal to 0)


so there is only one critical point at
. But this point does not fall in the region
. There are no extreme values in the region of interest, so we check the boundary.
Parameterize the boundary of
by


with
. Then
can be considered a function of
alone:



has critical points where
:



but
for all
, so this case yields nothing important.
At these critical points, we have temperatures of


so the plate is hottest at (1, 0) with a temperature of 14 (degrees?) and coldest at (-1, 0) with a temp of -12.