Question

Consider function 5E%7B2%7D%20%7D" id="TexFormula1" title="f(x) = 1+ \frac{5}{x} + \frac{2}{x^{2} }" alt="f(x) = 1+ \frac{5}{x} + \frac{2}{x^{2} }" align="absmiddle" class="latex-formula"> :
a) Find the interval where the function is decreasing. (Enter your answer using interval notation.)
b) Find the local maximum and minimum values.
c) Find the inflection points

Answer 1

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:

$f(x) = 1 + \frac{5}{x} + \frac{2}{x^2}$

a)$\frac{d}{dx}(f(x))= -\frac{5}{x^2} -\frac{2}{x^3}$

$\frac{d}{dx}(f(x))= 0$ 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) = $\frac{5}{x^3} +\frac{2}{x^4}$

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