Question

What is the average value of latex: y=\tan\left(\frac{x^2}{9}\right) y = tan ⁡ ( x 2 9 ) on the closed interval [1.25, 2]?

Answer 1

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