a. To evaluate the value of the temperature at t=5, we need to substitute t=5 into the given function and simplify. Using a calculator, we get:
O(5) = 300 - 100e^(-0.1*5) O(5) = 300 - 100e^(-0.5) O(5) = 300 - 60.65 O(5) = 239.35
Therefore, the value of the temperature at t=5 is 239.35°C, accurate to 2 d.p.
b. To find the value of O(t) for very large values of t, we need to consider the behavior of the exponential term as t approaches infinity. As t gets larger and larger, the exponent -0.1t gets more and more negative, which means that e^(-0.1t) gets closer and closer to zero. Therefore, the term 100e^(-0.1t) becomes negligible compared to the constant term 300. So, we can say that:
O(t) ≈ 300 - 0 O(t) ≈ 300
Therefore, the value of O(t) for very large values of t is approximately 300°C.
c. To find the inverse of the function O(t), we need to swap the roles of O and t and solve for O in terms of t. That is, we need to find a function e(t) such that e(O(t)) = t for any value of t. We can do this by following these steps:
Step 1: Rewrite O(t) as t and e(t) as O.
t = 300 - 100e^(-0.1O)
Step 2: Isolate the exponential term by subtracting 300 from both sides and multiplying by -1.
-100e^(-0.1O) = -t + 300
Step 3: Divide both sides by -100 to get rid of the coefficient.
e^(-0.1O) = (-t + 300)/-100
Step 4: Apply the natural logarithm function to both sides to get rid of the base e.
-0.1O = ln((-t + 300)/-100)
Step 5: Multiply both sides by -10 to get rid of the coefficient -0.1.
O = -10ln((-t + 300)/-100)
Therefore, the inverse of the function O(t) is e(t) = -10ln((-t + 300)/-100).