For Exercises 43-46, the vertices of a triangle are defined by the given points. To the nearest tenth, determine a. the perimeter of the triangle. b. the area of the triangle. c. the measure of the angles in the triangle. A 1 , 5 , B 5 , 8 , C 10 , 3
For Exercises 43-46, the vertices of a triangle are defined by the given points. To the nearest tenth, determine a. the perimeter of the triangle. b. the area of the triangle. c. the measure of the angles in the triangle. A 1 , 5 , B 5 , 8 , C 10 , 3
Solution Summary: The author calculates the perimeter of the triangle with points A, B, and C, using the equation of line joined by two points.
Given lim x-4 f (x) = 1,limx-49 (x) = 10, and lim→-4 h (x) = -7 use the limit properties
to find lim→-4
1
[2h (x) — h(x) + 7 f(x)] :
-
h(x)+7f(x)
3
O DNE
17. Suppose we know that the graph below is the graph of a solution to dy/dt = f(t).
(a) How much of the slope field can
you sketch from this information?
[Hint: Note that the differential
equation depends only on t.]
(b) What can you say about the solu-
tion with y(0) = 2? (For example,
can you sketch the graph of this so-
lution?)
y(0) = 1
y
AN
(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
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