I have a couple of questions and i wanted to know if my answers are correct. Thanks in advanced!
1. Determine the values of x for which the function f(x)=x+2/x-2 is discontinuous
Would the answers be x=2 because the function would be undefined?
2. An oil tank is being drained for cleaning. After t minutes there are v litres of oil left in the tank, where v(t)=35(25-t), 0<=t<=25
a) determine what v(25) represents
I substituted t+25 into the equation and got 0 as my answer. Would the 0 represent that the tank is completely drained?
b) determine the average rate of change of volume during the first 15 minutes
i got -1225 as my answer (would there be a unit, like litres?)
c) determine the rate of change of volume at the time t=15 minutes
is the question asking for the instantaneous rate of change? i'm not sure what they are asking for
your value x=2 is correct.
a) yes, but v(25) represents the amount of fluid remaining after 25 minutes. It just happens also to be true that at that time the tank is empty.
b)
v(0) = 875
v(15) = 350
volume changed by -525 in 15 minutes, for an average rate of change of
-525L/15min = -35 L/min
c) the rate of change is constant: -35L/min since
v(t) = -35t + 875
instantaneous rate of change is the derivative
v'(t) = -35
1. To determine the values of x for which the function f(x) = x+2/x-2 is discontinuous, we need to look for values of x that would make the denominator (x-2) equal to zero, because division by zero is undefined. So, to find those values, we solve the equation x-2=0, which gives x=2. So, yes, you're correct that x=2 would make the function undefined and discontinuous.
2. a) To determine what v(25) represents, we substitute t=25 into the expression v(t) = 35(25-t): v(25) = 35(25-25) = 35(0) = 0. So, your answer of 0 is correct. It means that after 25 minutes, there is no oil left in the tank, indicating that the tank is completely drained.
b) To determine the average rate of change of volume during the first 15 minutes, we need to find the change in volume over that time interval and divide it by the duration of the interval. By substituting t=0 and t=15 into the expression v(t) = 35(25-t), we can find the volumes at those times: v(0) = 35(25-0) = 35(25) = 875, and v(15) = 35(25-15) = 35(10) = 350. The change in volume over the first 15 minutes is v(15) - v(0) = 350 - 875 = -525. To find the average rate of change, we divide this change in volume by the duration of the interval, which is 15 minutes: -525 / 15 = -35. So, the average rate of change of volume during the first 15 minutes is -35 liters per minute.
c) To determine the rate of change of volume at the time t=15 minutes, we are indeed looking for the instantaneous rate of change of volume at that specific time. This is equivalent to finding the derivative of the volume function v(t) = 35(25-t) with respect to time t, and then evaluating it at t=15. Taking the derivative, we get dv/dt = -35. So, the rate of change of volume at the time t=15 minutes is -35 liters per minute.