A). √36 = +6 and -6 . Rational
B). √49 = +7 and -7 . Rational
C). √13 = +3.60555... and -3.60555... . Decimal never ends.
Irrational
D). √121 = +11 and -11 . Rational
Answer:
See picture above
Explanation:
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Answer:
The slope of the line is 0. A line with slope 0 represents a HORIZONTAL line.
Step-by-step explanation:
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<span>These are points where f ' = 0. Use the quiotent rule to find f '.
f ' (x) = [(x^3+2)(1) - (x)(3x^2)] / (x^3+2)^2
f ' (x) = (2 - 2x^3) / (x^3 + 2)^2
Set f ' (x) = 0 and solve for x.
f ' (x) = 0 = (2-2x^3) / (x^3+2)^2
Multiply both sides by (x^3+2)^2
(x^3+2)^2 * 0 = (x^3+2)^2 * [(2-2x^3)/(x^3+2)^2]
0 = 2 - 2x^3
Add 2x^3 to both sides
2x^3 + 0 = 2x^3 + 2 - 2x^3
2x^3 = 2
Divide both sides by 2
2x^3 / 2 = 2 / 2
x^3 = 1
Take cube roots of both sides
cube root (x^3) = cube root (1)
x = 1. This is our critical point
2) Points where f ' does not exist.
We know f ' (x) = (2-2x^3) / (x^3+2)^2
You cannot divide by 0 ever so f ' does not exist where the denominator equals 0
(x^3 + 2)^2 = 0. Take square roots of both sides
sqrt((x^3+2)^2) = sqrt(0)
x^3 + 2 = 0. Add -2 to both sides.
-2 + x^3 + 2 = -2 + 0
x^3 = -2. Take cube roots of both sides.
cube root (x^3) = cube root (-2)
x = cube root (-2). This is where f ' doesnt exist. However, it is not in our interval [0,2]. Thus, we can ignore this point.
3) End points of the domain.
The domain was clearly stated as [0, 2]. The end points are 0 and 2.
Therefore, our only options are: 0, 1, 2.
Check the intervals
[0, 1] and [1, 2]. Pick an x value in each interval and determine its sign.
In [0, 1]. Check 1/2. f ' (1/2) = (7/4) / (17/8)^2 which is positive.
In [1, 2]. Check 3/2. f ' (3/2) = (-19/4) / (43/8)^2 which is negative.
Therefore, f is increasing on [0, 1] and decreasing on [1, 2] and 1 is a local maximum.
f (0) = 0
f (1) = 1/3
f (2) = 1/5
Therefore, 0 is a local and absoulte minimum. 1 is a local and absolute
maximum. Finally, 2 is a local minimum. </span><span>Thunderclan89</span>
Answer:
hold on what is the problem?
Step-by-step explanation:
also 113
