Answer In a poligon of n sides can be drawn n-3 diagonals from each vertex.
Explanation The triangle is the first closed polygon, which has three sides and it is not possible to draw any diagonal from a vertex. In a polygon with four sides it is already possible to draw a diagonal from each vertex to the opposite one. Then diagonals = n-3.
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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
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(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, ∞)
