For this case we have the following function:
[tex]f(x) = tan(\frac{x^2}{9}) [/tex]
The average rate of change is given by:
[tex]AVR = \frac{f(x2) - f(x1)}{x2- x1} [/tex]
Evaluating the function for the given interval we have:
For x = 1.25:
[tex]f(1.25) = tan(\frac{1.25^2}{9}) [/tex]
[tex]f(1.25) = 0.18 [/tex]
For x = 2:
[tex]f(2) = tan(\frac{2^2}{9}) [/tex]
[tex]f(2) = 0.48 [/tex]
Then, replacing values we have:
[tex]AVR = \frac{0.48 - 0.18}{2 - 1.25} [/tex]
[tex]AVR = 0.4 [/tex]
Answer:
the average value of on the closed interval [1.25, 2] is:
[tex]AVR = 0.4 [/tex]
