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2 answers:
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Answer:
a) increasing (0, ∞); decreasing (-∞, 0)
b) min: (0, 0); max: DNE
c) inflection points: (-√6/3, 1/4), (√6/3, 1/4);
up: (-√6/3, √6/3); down: (-∞, -√6/3) ∪ (√6/3, ∞)
Step-by-step explanation:
The intervals of increase or decrease can be found from the sign of the slope, that is, the sign of the first derivative. That derivative is ...
f'(x) = ((x^2 +2)(2x) -x^2)(2x)/(x^2 +2)^2
f'(x) = 4x/(x^2 +2)^2
(a) f'(x) is positive for x > 0, hence ...
the function is increasing on (0, ∞)
the function is decreasing on (-∞, 0)
__
(b) f'(x) is zero for x=0, a local minimum. f(0) = 0.
minimum: (0, 0)
maximum: DNE
__
(c) The second derivative is ...
f''(x) = ((x^2+2)^2·4 -(4x)(2)(x^2 +2)(2x))/(x^2 +2)^4
= (8 -12x^2)/(x^2 +2)^3
Inflection points are where the second derivative is zero, or ...
8 -12x^2 = 0
x^2 = 2/3
x = ±√(2/3) = ±(√6)/3
The values of f(x) there are x^2/(x^2 +2) = (2/3)/(2/3 +2) = (2/8) = 1/4
The points of inflection are (x, y) = (-√6/3, 1/4), (√6/3, 1/4).
The function is concave up between these inflection points
f(x) is concave up on the interval (-√6/3, √6/3)
f(x) is concave down on (-∞, -√6/3) ∪ (√6/3, ∞)
To solve this problem, we have to consider the fact that if the probability that the cyclist may crash due to some circumstance. The probability the cyclist doesn't crash at any given day is 3/8
<h3>Probability</h3>The probability the cyclist crash when it's wet or dry will be the
- P(wet) = 5/8
- P(dry = 1/8
The probability the cyclist crash on any given day is
we can subtract 1 from the value above to find the probability the cyclist doesn't crash at any given day
The probability the cyclist doesn't crash at any given day is 3/8
Learn more on probability here;
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Answer:
A. m =
Step-by-step explanation:
<em><u>The variables "ma" are connected, implying they are together through multiplication. To remove "a" from "m," you must do the opposite, which is division. So, divide both sides by "a," like so:</u></em>
F = ma
= m