anonymous
  • anonymous
Determine algebraically whether the function is even, odd, or neither even nor odd. Odd Even Neither
Mathematics
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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anonymous
  • anonymous
Determine algebraically whether the function is even, odd, or neither even nor odd. Odd Even Neither
Mathematics
schrodinger
  • schrodinger
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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anonymous
  • anonymous
function is \[f(x)=x+\frac{ 12 }{ x }\]
hartnn
  • hartnn
Find f(-x) by replacing x in f(x) by -x. If f(-x) = f(x), then the function is even If f(-x) = -f(x), then the function is odd else neither.
anonymous
  • anonymous
\[f(-x)=-x-\frac{12}{x}=-(x+\frac{12}{x})=-f(x)\]

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anonymous
  • anonymous
so is an odd function?
anonymous
  • anonymous
yes it is
anonymous
  • anonymous
how about \[f(x)=x+x^2\]
anonymous
  • anonymous
thanks
anonymous
  • anonymous
no problem :)
anonymous
  • anonymous
neither?
anonymous
  • anonymous
yes
anonymous
  • anonymous
yey!! thank you

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