Assume (a,b) has a minimum element m.
m is in the interval so a < m < b.
a < m
Adding a to both sides,
2a < a + m
Adding m to both sides of the first inequality,
a + m < 2m
So
2a < a+m < 2m
a < (a+m)/2 < m < b
Since the average (a+m)/2 is in the range (a,b) and less than m, that contradicts our assumption that m is the minimum. So we conclude there is no minimum since given any purported minimum we can always compute something smaller in the range.
Answer:
Analyzed and Sketched.
Step-by-step explanation:
We are given 
To sketch the graph we need to find 2 components.
1) First derivative of y with respect to x to determine the interval where function increases and decreases.
2) Second derivative of y with respect to x to determine the interval where function is concave up and concave down.

is absolute maximum

is the point concavity changes from down to up.
Here, x = 0 is vertical asymptote and y = 0 is horizontal asymptote.
The graph is given in the attachment.
Answer:
1. The independent variable goes on the x-axis and the dependent variable goes on the y-axis.