Answer:
See below
Step-by-step explanation:
I assume the function is ![f(x)=1+\frac{5}{x}-\frac{4}{x^2}](https://tex.z-dn.net/?f=f%28x%29%3D1%2B%5Cfrac%7B5%7D%7Bx%7D-%5Cfrac%7B4%7D%7Bx%5E2%7D)
A) The vertical asymptotes are located where the denominator is equal to 0. Therefore,
is the only vertical asymptote.
B) Set the first derivative equal to 0 and solve:
![f(x)=1+\frac{5}{x}-\frac{4}{x^2}](https://tex.z-dn.net/?f=f%28x%29%3D1%2B%5Cfrac%7B5%7D%7Bx%7D-%5Cfrac%7B4%7D%7Bx%5E2%7D)
![f'(x)=-\frac{5}{x^2}+\frac{8}{x^3}](https://tex.z-dn.net/?f=f%27%28x%29%3D-%5Cfrac%7B5%7D%7Bx%5E2%7D%2B%5Cfrac%7B8%7D%7Bx%5E3%7D)
![0=-\frac{5}{x^2}+\frac{8}{x^3}](https://tex.z-dn.net/?f=0%3D-%5Cfrac%7B5%7D%7Bx%5E2%7D%2B%5Cfrac%7B8%7D%7Bx%5E3%7D)
![0=-5x+8](https://tex.z-dn.net/?f=0%3D-5x%2B8)
![5x=8](https://tex.z-dn.net/?f=5x%3D8)
![x=\frac{8}{5}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B8%7D%7B5%7D)
Now we test where the function is increasing and decreasing on each side. I will use 2 and 1 to test this:
![f'(2)=-\frac{5}{2^2}+\frac{8}{2^3}=-\frac{5}{4}+\frac{8}{8}=-\frac{5}{4}+1=-\frac{1}{4}](https://tex.z-dn.net/?f=f%27%282%29%3D-%5Cfrac%7B5%7D%7B2%5E2%7D%2B%5Cfrac%7B8%7D%7B2%5E3%7D%3D-%5Cfrac%7B5%7D%7B4%7D%2B%5Cfrac%7B8%7D%7B8%7D%3D-%5Cfrac%7B5%7D%7B4%7D%2B1%3D-%5Cfrac%7B1%7D%7B4%7D)
![f'(1)=-\frac{5}{1^2}+\frac{8}{1^3}=-\frac{5}{1}+\frac{8}{1}=-5+8=3](https://tex.z-dn.net/?f=f%27%281%29%3D-%5Cfrac%7B5%7D%7B1%5E2%7D%2B%5Cfrac%7B8%7D%7B1%5E3%7D%3D-%5Cfrac%7B5%7D%7B1%7D%2B%5Cfrac%7B8%7D%7B1%7D%3D-5%2B8%3D3)
Therefore, the function increases on the interval
and decreases on the interval
.
C) Since we determined that the slope is 0 when
from the first derivative, plugging it into the original function tells us where the extrema are. Therefore,
, meaning there's an extreme at the point
, but is it a maximum or minimum? To answer that, we will plug in
into the second derivative which is
. If
, then it's a minimum. If
, then it's a maximum. If
, the test fails. So,
, which means
is a local maximum.
D) Now set the second derivative equal to 0 and solve:
![f''(x)=\frac{10}{x^3}-\frac{24}{x^4}](https://tex.z-dn.net/?f=f%27%27%28x%29%3D%5Cfrac%7B10%7D%7Bx%5E3%7D-%5Cfrac%7B24%7D%7Bx%5E4%7D)
![0=\frac{10}{x^3}-\frac{24}{x^4}](https://tex.z-dn.net/?f=0%3D%5Cfrac%7B10%7D%7Bx%5E3%7D-%5Cfrac%7B24%7D%7Bx%5E4%7D)
![0=10x-24](https://tex.z-dn.net/?f=0%3D10x-24)
![-10x=-24](https://tex.z-dn.net/?f=-10x%3D-24)
![x=\frac{24}{10}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B24%7D%7B10%7D)
![x=\frac{12}{5}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B12%7D%7B5%7D)
We then test where
is negative or positive by plugging in test values. I will use -1 and 3 to test this:
, so the function is concave down on the interval ![(-\infty,0)\cup(0,\frac{12}{5})](https://tex.z-dn.net/?f=%28-%5Cinfty%2C0%29%5Ccup%280%2C%5Cfrac%7B12%7D%7B5%7D%29)
, so the function is concave up on the interval ![(\frac{12}{5},\infty)](https://tex.z-dn.net/?f=%28%5Cfrac%7B12%7D%7B5%7D%2C%5Cinfty%29)
The inflection point is where concavity changes, which can be determined by plugging in
into the original function, which would be
, or
.
E) See attached graph