findind the absolute
- Thread starter dear2009
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dear2009 said:
First plot your function and observe the absolute minimum within the given domain.
How much is it? where (x) is it?
Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.
BigGlenntheHeavy
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\(\displaystyle f(x) \ = \ x+1+\frac{5}{x+2}, \ f \ ' \ (x) \ = \ 1-\frac{5}{(x+2)^{2}} \ = \ \frac{x^{2}+4x-1}{(x+2)^{2}}\)
\(\displaystyle Setting \ the \ slope \ = \ to \ zero, \ we \ have \ x \ = \ \sqrt5-2, \ over \ the \ interval \ [0,1].\)
\(\displaystyle Hence, \ f(\sqrt5-2) \ = \ 2\sqrt5-1, \ thus \ min. \ point \ = \ (\sqrt5-2,2\sqrt5-1).\)
\(\displaystyle Question: \ What \ are \ the \ equations \ for \ the \ two \ asymptotes \ of \ f(x) \ = \ x+1+\frac{5}{x+2}?\)
see graph below:
[attachment=0:n568zmeq]xxx.jpg[/attachment:n568zmeq]
\(\displaystyle Setting \ the \ slope \ = \ to \ zero, \ we \ have \ x \ = \ \sqrt5-2, \ over \ the \ interval \ [0,1].\)
\(\displaystyle Hence, \ f(\sqrt5-2) \ = \ 2\sqrt5-1, \ thus \ min. \ point \ = \ (\sqrt5-2,2\sqrt5-1).\)
\(\displaystyle Question: \ What \ are \ the \ equations \ for \ the \ two \ asymptotes \ of \ f(x) \ = \ x+1+\frac{5}{x+2}?\)
see graph below:
[attachment=0:n568zmeq]xxx.jpg[/attachment:n568zmeq]
