Chapter 1.2, Problem 18E

(b) Find the (instantaneous) rate of change of y at x = 5. In the previous part, we found the average rate of change for several intervals of decreasing size starting at x = 5. The instantaneous rate of change of fat x = 5 is the limit of the average rate of change over the interval [x, x + h] as h approaches 0. This is given by the derivative in the following limit. lim h→0 - f(x + h) − f(x) h The first step to find this limit is to compute f(x + h). Recall that this means replacing the input variable x with the expression x + h in the rule defining f. f(x + h) = (x + h)² - 5(x+ h) = 2xh+h2_ x² + 2xh + h² 5✔ - 5 )x - 5h Step 4 - The second step for finding the derivative of fat x is to find the difference f(x + h) − f(x). - f(x + h) f(x) = = (x² x² + 2xh + h² - ])- = 2x + h² - 5h ])x-5h) - (x² - 5x) = ]) (2x + h - 5) Macbook Pro

Evaluate the integral using integration by parts. Sx² cos (9x) dx

Let f be defined as follows. y = f(x) = x² - 5x (a) Find the average rate of change of y with respect to x in the following intervals. from x = 4 to x = 5 from x = 4 to x = 4.5 from x = 4 to x = 4.1 (b) Find the (instantaneous) rate of change of y at x = 4. Need Help? Read It Master It