determine intervals , minima , maxima,...?

f(x) = x^2 / (x-4)

f'(x) = [(x-4)(2x) - x^2(1)] /(x-4)^2 (using quotient rule)

f'(x) = (2x^2-8x-x^2) / (x-4)^2

f'(x) = (x^2-8x) /(x-4)^2 =0

x^2-8x=0

x(x-8)=0

x=0 and x=8 are extreme points

f''(x) = [(x-4)^2 (2x-8) - (x^2-8x) (2)(x-4) ] / (x-4)^4

f''(x) = [ 2(x-4)^3 - 2x(x-4) (x-8) ] / (x-4)^4

when x=0

f''(x) = -0.5 < 0

Therefore, f has a maximum at x=0

when x=8,

f''(x)=0.5 > 0

Therefore, f has a minimum at x=8

Relative minimum occurs at (8, f(8)) = (8,16)

Relative maximum occurs at (0,0)

To find the intervals where f is increasing and decreasing:

f'(x) =0 gives a solution of x=0, x=8

Consider the following intervals:

(-∞,0)(0,8),(8,∞)

Choose any point from each interval and examine the sign of f'(x).

f'(x)= (x^2-8x) /(x-4)^2

(-∞,0) : choose -5 : f'(-5) = 0.802 >0 : f is increasing

(0,8) : choose 5: f'(5)=-15 < 0 : f is decreasing

(8,∞) : choose x=10 : f'(10)= 0.5 > 0 : f is increasing.

Therefore,

f(x) is increasing on (-∞,0) and (8,∞)

f(x) is decreasing on (0,8)

Note:

The function is discontinuous at x=4;

The extreme factors are as you state, as are whilst the function is reducing and increasing. The community minima are as you assert. i'd call the factors at x=-3 and +3 "absolute maxima" considering community max/min are many times the place f'(x)=0. Absolute factors are often observed as "worldwide" which in line with threat describe the factors extra effective. notice that f(x) = (a million/4)(x^2 - 4)^2 + 6 so x=-2 and +2 are absolute(worldwide) minima additionally.

Find when the derivative is increasing or decreasing, by finding where f'(x) is zero. Between these zeros of f'(x), the value will be positive for f(x) increasing, or negative for f(x) decreasing