The oxygen content t days after organic waste was dumped into a pond is given by f (t) = 100 (t2 + 10t + 100/t2 + 20t + 100).

Question

A) use the intermediate value theorem to conclude that the oxygen content of the pond must have been at a level of 85 % at some time.

B) At what time(s) is the oxygen content at the 85% level

Answer 1

A) First, let's check continuity. The function is a rational function and is continuous for all t where the denominator is not zero. Since t²+20t+100 is always positive (it has no real roots and is a parabola opening upwards), f(t) is continuous for all t.

Then, we determine values at specific points:

f(0) = 100

f(10) = 75

Now, we apply the IVT:

From t=0 to t=10, f(t) moves from 100 to 75. Since 85 is between 100 and 75, by the IVT, there must be some c in the interval (0,10) such that f(c)= 85.

B) Set f(t) = 85, rearrange the quadratic such that zero is left along on the right-hand side, then use the quadratic formula to solve for t:

t = 2.25 days and 44.41 days

Thus, the oxygen content of the pond is at 85% at approximately t = 25 days and t = 44.41 days after the waste was dumped.

Hope this was helpful.