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Patrick made tortilla soup in his slow cooker and has turned the heat off to allow the soup to cool slightly before consuming. The temperature, in degrees Fahrenheit, of the soup x minutes after the heat was turned off is represented by the function below.

T(x) = 15 + 155e-0.02x

What is the approximate average rate of change over the interval [10, 30]?

A.
-6.05 degrees per minute
B.
-1.05 degrees per minute
C.
-41.84 degrees per minute
D.
-2.09 degrees per minute

User Amit Bakle
by
5.6k points

2 Answers

4 votes

Final answer:

To determine the average rate of change of the function T(x) over the interval [10, 30], calculate the function values at the endpoints, subtract them, and divide by the change in x. The average rate of change is approximately -6.05 degrees per minute, making the correct answer Option A.

Step-by-step explanation:

To find the average rate of change of the function T(x) = 15 + 155e-0.02x over the interval [10, 30], we need to calculate the difference in function values at these points and divide it by the difference in x-values.

  • First, calculate T(10) and T(30).
  • Then find the difference: T(30) - T(10).
  • Finally, divide by the change in x: (T(30) - T(10))/(30 - 10).

Here are the calculations:

T(10) = 15 + 155e-0.02(10) = 15 + 155e-0.2

T(30) = 15 + 155e-0.02(30) = 15 + 155e-0.6

Now subtract the lower x-value function from the higher, and divide by the number of units between the two x-values:

Average Rate of Change = (T(30) - T(10)) / (30 - 10)

Plugging in the values we get:

Average Rate of Change = ((15 + 155e-0.6) - (15 + 155e-0.2)) / 20

Calculating the numerical values (using a calculator):

Average Rate of Change ≈ (-6.05) degrees per minute

Therefore, the correct answer is Option A, which is approximately -6.05 degrees per minute.

User Greg Moens
by
6.3k points
6 votes

9514 1404 393

Answer:

D. -2.09 degrees per minute

Step-by-step explanation:

The average rate of change is the change in temperature divided by the change in time:

m = (T(30) -T(10))/(30 -10)

= ((15 +155 e^(-30·0.02)) -(15 +155 e^(-10·0.02))/20

= (155/20)(e^-0.6 -e^-0.2) ≈ -2.09 . . . degrees per minute

User Konstantin Samarin
by
5.8k points