Answer :
Answer:
Required (a) vertical tangent at (1,0) (b) horizontal tangent at (-1,2) (c) concave upward ob (1,0) (d) concave downward on (-1,2).
Step-by-step explanation:
Consider a parametric curve,
[tex]x=t, y=t+\frac{1}{t}[/tex]
Then substitute value of x in y we get,
[tex]y=x+\frac{1}{x}=f(x)[/tex]
(a) To find points where vertical or horizintal tangent are meet we have to take,
[tex]f'(x)=0\implies 1-\frac{1}{x^2}=0\implies x^2=1\implies x=\pm 1[/tex]
when,
x=1, y=0
x=-1, y=2
Now at point (1,0),
[tex]\lim_{(x,y)\to (1,0)}f(x)\to \infty[/tex]
So at this point there is a vertical tangent.
(b) And (-1,2) is then the point where horizontal tangent meet.
To find point of interval where function is concave upward or downward we have to derivative f(x) with respect to x twice that is,
[tex]f''(x)=1+ \frac{2}{x^3}[/tex]
(c) At x=1, f''(1)=3>0, that means on the interval (1,0) the function is concave upward.
(d) At x=-1, f''(-1)=-1, which imply on the interval (-1,2) the funcrion is concave downward.