Question

Show that \(({x^3} + 2x)/(2x + 1)\) is \(O({x^2})\).

Short answer

The given function\(({x^3} + 2x)/(2x + 1)\)is\(O({x^2})\)when\(k = 2\)and\(C = \frac{3}{2}\).

Step-by-step solution

Step 1:

First we find the quotient using the long division method.

\(2x + 1\mathop{\left){\vphantom{1\begin{array}{l}{x^3} + 2x\\\underline { - {x^3} - \frac{1}{2}{x^2}} \\0 - \frac{1}{2}{x^2} + 2x\\\underline {0 + \frac{1}{2}{x^2} + \frac{1}{4}x} \\0 + 0 + \frac{9}{4}x\\\underline {0 + 0 - \frac{9}{4}x - \frac{9}{8}} \\0 + 0 + 0 - \frac{9}{8}\end{array}}}\right.

\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}{x^3} + 2x\\\underline { - {x^3} - \frac{1}{2}{x^2}} \\0 - \frac{1}{2}{x^2} + 2x\\\underline {0 + \frac{1}{2}{x^2} + \frac{1}{4}x} \\0 + 0 + \frac{9}{4}x\\\underline {0 + 0 - \frac{9}{4}x - \frac{9}{8}} \\0 + 0 + 0 - \frac{9}{8}\end{array}}}}

\limits^{\displaystyle\,\,\, {\frac{1}{2}{x^2} - \frac{1}{3}x + \frac{2}{8}}}\)

We get,

\(\begin{array}{l}Q = \frac{1}{2}{x^2} - \frac{1}{3}x + \frac{2}{8}\\R = - \frac{9}{8}\end{array}\)

Where,

\(Q = \)Quotient

\(R = \)Remainder`

Step 2:

We can now write the fraction as,

\(\begin{array}{l}f(x) = \frac{{{x^3} + 2x}}{{2x + 1}}\\ = \left( {\frac{1}{2}{x^2} - \frac{1}{3}x + \frac{2}{8}} \right) - \frac{{\frac{9}{8}}}{{2x + 1}}\end{array}\)

When\(x > 2\)we get the property\({x^2} > x > \frac{9}{8}\).

When \(x > 0\), we get the property \(\frac{1}{4}x > 0\) and also \(\frac{{\frac{9}{8}}}{{2x + 1}} > 0\).

Step 3:

We will take\(k = 2\)and get\(x > 2\)

\(\begin{array}{l}\left| {f(x)} \right| = \left| {(\frac{1}{2}{x^2} - \frac{1}{4}x + \frac{9}{8}) - \frac{{\frac{9}{8}}}{{(2x + 1)}}} \right|\\ \le \left| {\frac{1}{2}{x^2} - \frac{1}{4}x + \frac{9}{8}} \right|\\ \le \left| {\frac{1}{2}{x^2} + \frac{9}{8}} \right|\\ \le \left| {\frac{1}{2}{x^2}} \right| + \left| {\frac{9}{8}} \right|\end{array}\)

\(\begin{array}{l} = 0 \le \frac{1}{2}{x^2} + \frac{9}{8}\\ \le \frac{1}{2}{x^2} + {x^2}\\ = \frac{3}{2}\left| {{x^2}} \right|\end{array}\)

Therefore taking\(C = \frac{3}{2}\)by the definition we get that\(({x^3} + 2x)/(2x + 1)\)is\(O({x^2})\)when \(k = 2\) and \(C = \frac{3}{2}\).