Question

Consider the equation below. f(x) = x^2/(x^2 + 2) (a) Find the interval on which f is increasing. (Enter your answer using interval notation.) Find the interval on which f is decreasing. (Enter your answer using interval notation.) (b) Find the local minimum and maximum values of f. (If an answer does not exist, enter DNE.) local minimum value local maximum value (c) Find the inflection points. (x, y) = (smaller x-value) (x, y) = (larger x-value) Find the interval on which f is concave up. (Enter your answer using interval notation.) Find the interval on which f is concave down. (Enter your answer using interval notation.)

Answer 1

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, ∞)