Answer:
D
Step-by-step explanation:
The average rate of change of f(x) in the closed interval [ a, b ] is
Here [ a, b ] = [ 2, 4 ]
From the table
f(b) = f(4) = 16
f(a) = f(2) = 4, hence
average rate of change = =
= 6
In most linear programming problems, there are two stages:
1 answer:
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The rate of change of temperature with time at a point in time is given by
the derivative of the function for the temperature of the soup.
The correct responses are;
- a) H'(5) is approximately<u> -2.6 degrees Celsius per minute</u>.
- b) Yes
- c) The equation for the line tangent is<u> y = -3.6·t + 90.8</u>
- The approximate value of C(5) is <u>72.8 °C</u>
- d) The rate of change of the temperature of the soup at t = 3 minutes is <u>-3.6 degrees Celsius per minute</u>.
Reasons:
a) From the data in the table, we have;
The approximate value of H'(5) is given by the average value of the rate of
change of the temperature with time between points, t = 3, and t = 8
Therefore;
Which gives;
Therefore, H'(5) = <u>-2.6°C per minute</u>
b) Given that the function is twice differentiable over the interval, 0 ≤ t ≤ 12, the function for the change in temperature is continuous in the interval 0 ≤ t ≤ 12
At t = 0, H(0) = 90 °C
At t = 12, H(12) = 58 °C
- 58 °C < 60 < 90 °C
Therefore, there exist a temperature, of 60 °C between 90° C and 58 °C
c) The given derivative of <em>C</em> is,
At t = 3, we have;
Therefore, we have;
y - 80 ≈ -3.1 × (x - 3)
The equation for the tangent is; y = -3.6 × (x - 3) + 80
y = -3.6·x + 10.8 + 80 = -3.6·x + 90.8
- The equation for the tangent is; <u>y = -3.6·x + 90.8</u>
The value of C(5) is approximately, C(5) ≈ -3.6 × 5 + 90.8 = 72.8
- <u>C(5) ≈ 72.8°</u>
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d) Based on the the model above, the rate at which the temperature of the
soup is changing at t = 3 minutes is <u>-3.6 degrees per minute</u>.
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