Whether the intermediate value theorem guarantees or not that the function Y 1 = 21 x 4 + 46 x 3 − 238 x 2 − 506 x + 77 has a zero in the interval [ − 4 , − 3 ] . The table is as follows: x Y 1 − 4 725 − 3 − 88 − 2 105 − 1 320 0 77 1 − 600 2 − 1183 3 − 640 4 2565 5 10472 6 25625
Whether the intermediate value theorem guarantees or not that the function Y 1 = 21 x 4 + 46 x 3 − 238 x 2 − 506 x + 77 has a zero in the interval [ − 4 , − 3 ] . The table is as follows: x Y 1 − 4 725 − 3 − 88 − 2 105 − 1 320 0 77 1 − 600 2 − 1183 3 − 640 4 2565 5 10472 6 25625
Solution Summary: The author evaluates whether the intermediate value theorem guarantees or not that the function Y_1=21x4+46
Whether the intermediate value theorem guarantees or not that the function Y1=21x4+46x3−238x2−506x+77 has a zero in the interval [−4,−3]. The table is as follows:
x
Y1
−4
725
−3
−88
−2
105
−1
320
0
77
1
−600
2
−1183
3
−640
4
2565
5
10472
6
25625
(b)
To determine
Whether the intermediate value theorem guarantees or not that the function Y1=21x4+46x3−238x2−506x+77 has a zero in the interval [−3,−2]. The table is as follows:
x
Y1
−4
725
−3
−88
−2
105
−1
320
0
77
1
−600
2
−1183
3
−640
4
2565
5
10472
6
25625
(c)
To determine
Whether the intermediate value theorem guarantees or not that the function Y1=21x4+46x3−238x2−506x+77 has a zero in the interval [−2,−1]. The table is as follows:
x
Y1
−4
725
−3
−88
−2
105
−1
320
0
77
1
−600
2
−1183
3
−640
4
2565
5
10472
6
25625
(d)
To determine
Whether the intermediate value theorem guarantees or not that the function Y1=21x4+46x3−238x2−506x+77 has a zero in the interval [−1,0]. The table is as follows:
Suppose f and g are the piecewise-defined functions defined
here. For each combination of functions in Exercises 51–56,
(a) find its values at x = -1, x = 0, x = 1, x = 2, and x = 3,
(b) sketch its graph, and (c) write the combination as a
piecewise-defined function.
f(x) = {
(2x + 1, ifx 0
g(x) = {
-x, if x 2
8(4):
51. (f+g)(x)
52. 3f(x)
53. (gof)(x)
56. g(3x)
54. f(x) – 1
55. f(x – 1)
For Exercises 103–104, given y = f(x),
remainder
a. Divide the numerator by the denominator to write f(x) in the form f(x) = quotient +
divisor
b. Use transformations of y
1
to graph the function.
2x + 7
5х + 11
103. f(x)
104. f(x)
x + 3
x + 2
In Exercises 83–85, you will use a CAS to help find the absolute extrema of the given function over the specified closed interval. Per-form the following steps.
a. Plot the function over the interval to see its general behavior there.
b. Find the interior points where ƒ′ = 0. (In some exercises, you may have to use the numerical equation solver to ap-proximate a solution.) You may want to plot ƒ′ as well.
c. Find the interior points where ƒ′ does not exist.
d. Evaluate the function at all points found in parts (b) and (c) and at the endpoints of the interval.
e. Find the function’s absolute extreme values on the interval and identify where they occur.
83. ƒ(x) = x4 - 8x2 + 4x + 2, [-20/25, 64/25]
84. ƒ(x) = -x4 + 4x3 - 4x + 1, [-3/4, 3] 85. ƒ(x) = x^(2/3)(3 - x), [-2, 2]
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