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## Here's the Solution to this Question

$f(n)=\dfrac{n^2+1}{n+1}\leq\dfrac{n^2+n}{n+1}=n, n\in \N$

$\N=\{1,2,...\}$

By the definition $f(n)$ is $O(n)$ with $C=1$ and $k=1.$

$\dfrac{n^2+1}{n+1}-\dfrac{n}{2}=\dfrac{2n^2+2-n^2-n}{2(n+1)}=$

$=\dfrac{n^2-n+\dfrac{1}{4}+\dfrac{7}{4}}{2(n+1)}=\dfrac{(n-\dfrac{1}{2})^2+\dfrac{7}{4}}{2(n+1)}>0,n\in\N$

$|f(n|)=|\dfrac{n^2+1}{n+1}|>\dfrac{1}{2} |n|, n\in\N$

Then $f(n)$ is $\Omega(n).$

$\dfrac{1}{2}|n|<|f(n)|\leq|n|, n\in\N$

By the definition this means that $f(n)=\dfrac{n^2+1}{n+1}$ is $\Theta(n), n\in\N.$

$f(n)=\dfrac{n^2+1}{n+1}\leq\dfrac{n^2+n}{n+1}=n, n\in \N$

$\N=\{1,2,...\}$

By the definition $f(n)$ is $O(n)$ with $C=1$ and $k=1.$

$\dfrac{n^2+1}{n+1}-\dfrac{n}{2}=\dfrac{2n^2+2-n^2-n}{2(n+1)}=$

$=\dfrac{n^2-n+\dfrac{1}{4}+\dfrac{7}{4}}{2(n+1)}=\dfrac{(n-\dfrac{1}{2})^2+\dfrac{7}{4}}{2(n+1)}>0,n\in\N$

$|f(n|)=|\dfrac{n^2+1}{n+1}|>\dfrac{1}{2} |n|, n\in\N$

Then $f(n)$ is $\Omega(n).$

$\dfrac{1}{2}|n|<|f(n)|\leq|n|, n\in\N$

By the definition this means that $f(n)=\dfrac{n^2+1}{n+1}$ is $\Theta(n), n\in\N.$

$f(n)=\dfrac{n^2+1}{n+1}\leq\dfrac{n^2+n}{n+1}=n, n\in \N$

$\N=\{1,2,...\}$

By the definition $f(n)$ is $O(n)$ with $C=1$ and $k=1.$

$\dfrac{n^2+1}{n+1}-\dfrac{n}{2}=\dfrac{2n^2+2-n^2-n}{2(n+1)}=$

$=\dfrac{n^2-n+\dfrac{1}{4}+\dfrac{7}{4}}{2(n+1)}=\dfrac{(n-\dfrac{1}{2})^2+\dfrac{7}{4}}{2(n+1)}>0,n\in\N$

$|f(n|)=|\dfrac{n^2+1}{n+1}|>\dfrac{1}{2} |n|, n\in\N$

Then $f(n)$ is $\Omega(n).$

$\dfrac{1}{2}|n|<|f(n)|\leq|n|, n\in\N$

By the definition this means that $f(n)=\dfrac{n^2+1}{n+1}$ is $\Theta(n), n\in\N.$

$f(n)=\dfrac{n^2+1}{n+1}\leq\dfrac{n^2+n}{n+1}=n, n\in \N$

$\N=\{1,2,...\}$

By the definition $f(n)$ is $O(n)$ with $C=1$ and $k=1.$

$\dfrac{n^2+1}{n+1}-\dfrac{n}{2}=\dfrac{2n^2+2-n^2-n}{2(n+1)}=$

$=\dfrac{n^2-n+\dfrac{1}{4}+\dfrac{7}{4}}{2(n+1)}=\dfrac{(n-\dfrac{1}{2})^2+\dfrac{7}{4}}{2(n+1)}>0,n\in\N$

$|f(n|)=|\dfrac{n^2+1}{n+1}|>\dfrac{1}{2} |n|, n\in\N$

Then $f(n)$ is $\Omega(n).$

$\dfrac{1}{2}|n|<|f(n)|\leq|n|, n\in\N$

By the definition this means that $f(n)=\dfrac{n^2+1}{n+1}$ is $\Theta(n), n\in\N.$