Let ∑ u k be a series and define series ∑ p k and ∑ q k so that p k = 0 , u k , u k ≤ 0 u k > 0 and q k = − u k , 0 , u k < 0 u k ≥ 0 (a) Show that ∑ u k converges absolutely if and only if ∑ p k and ∑ q k both coverage. (b) Show that if one of ∑ p k or ∑ q k converges and the other diverges, then ∑ u k diverges. (c) Show that if ∑ u k converges conditionally, then both ∑ p k or ∑ q k diverge.
Let ∑ u k be a series and define series ∑ p k and ∑ q k so that p k = 0 , u k , u k ≤ 0 u k > 0 and q k = − u k , 0 , u k < 0 u k ≥ 0 (a) Show that ∑ u k converges absolutely if and only if ∑ p k and ∑ q k both coverage. (b) Show that if one of ∑ p k or ∑ q k converges and the other diverges, then ∑ u k diverges. (c) Show that if ∑ u k converges conditionally, then both ∑ p k or ∑ q k diverge.
(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.
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