Answer:
a) x<0.4 and x>0
b) (0.4, -1/8) is a relative min. there is no relative max
c)inflection at (0.4, -1/8)
Step-by-step explanation:
[tex]f(x) = 1 + \frac{5}{x} + \frac{2}{x^2}[/tex]
a)[tex]\frac{d}{dx}(f(x))= -\frac{5}{x^2} -\frac{2}{x^3}[/tex]
[tex]\frac{d}{dx}(f(x))= 0[/tex] when x = -0.4
looking at the graph of f'(x), the value of f'(x) is only positive for points in the set -0.4<x<0, which means f(x) is decreasing for intervals x<0.4 and x>0
I don't know interval notation, none of my math classes or teachers have ever required me or taught me how to use interval notation so yeah
b) we see that f'(x) = 0 when x = 0.4 and f'(x) is undefined at x = 0
0.4 and 0 are candidates for our max or min
f(0.4) = -1/8
f(0) is nonexistent, so it will not be our max or min
at 0.4 f'(x) is negative to left and positive to right, so (0.4, -1/8) is a relative min. there is no max
c) inflection where f''(x) is 0 or undefined
d''(x) = [tex]\frac{5}{x^3} +\frac{2}{x^4}[/tex]
0 when x = 0.4
undefined at x = 0
f''(x) goes from negative to positive when x = 0.4
reflection at (0.4, -1/8)
f(0) is undefined, so there is no inflection when x = 0
