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Another Example

Example 2: Let tex2html_wrap_inline18 Graph f.

(1) Vertical Asymptote(s). By inspection, x=0 is the vertical asymptote.

(a) As tex2html_wrap_inline24 , say x=0.0001, we get f(0.0001)= tex2html_wrap_inline30

(b) As tex2html_wrap_inline32 say x=-0.0001, we get tex2html_wrap_inline36

(2) Horizontal Asymptote: y=2

As tex2html_wrap_inline108, say x=103, tex2html_wrap_inline44 and as tex2html _wrap_inline114 say x=-103, tex2html_wrap_inline50 Therefore the horizontal asymtote for f is y=2.

(3) Find the interval(s) where f is increasing or decreasing.

tex2html_wrap_inline58 .

Thus, f is decreasing in tex2html_wrap_inline140 0) and increasing in tex2html_wrap_inline68 and f has no maximum or minimum since x=0 is undefined.

(4) Find the interval where f is concave upward or downward. Since tex2html_wrap_inline76 tex2html_wrap_inline78 f is alw ays concave downward. There is no inflection point since x=0 is undefined.

(5) Use MFI to sketch the graph of f.