- Find and simplify the coefficient of \(x^7\) in the binomial expansion of \(\left( x^2 + \dfrac{2}{x} \right)^8\).
The
binomial theorem says that, if
\(n\) is a positive integer, then
\[\begin{equation*}
(x+y)^n = \sum_{r=0}^n \binom{n}{r} x^r y^{n-r}
\end{equation*}\]
where
\[\begin{equation*}
\binom{n}{r} = \frac{n!}{r!(n-r)!}
\end{equation*}\]
So \(\left(x^2+\dfrac{2}{x}\right)^8\) is made up of terms of the form \(\dbinom{8}{r}x^{2r}\left(\dfrac{2}{x}\right)^{8-r}\).
To find the value of \(r\) that gives the term in \(x^7\), we require \(2r-(8-r)=7\), which gives \(r = 5\).
So the coefficient of \(x^7\) is \(\dbinom{8}{5}2^3\), or \(448\).
- Calculate the minimum value of \(x^2 + \dfrac{2}{x}\) for positive values of \(x\).
If we let
\[\begin{equation*}
y = x^2 + \frac{2}{x} = x^2 + 2x^{-1}
\end{equation*}\]
then we have that
\[\begin{equation*}
\frac{dy}{dx} = 2x - 2x^{-2}.
\end{equation*}\]
Hence,
\[\begin{equation*}
\frac{dy}{dx} = 0 \iff x = x^{-2} \iff x^3 = 1 \iff x = 1.
\end{equation*}\]
Is this stationary point a minimum? Yes: as \(x \to 0\) or \(x \to \infty\), \(y \to \infty\), so this stationary point has to be a minimum.
The minimum value of the expression is then \(1^2+\dfrac{2}{1} = 3\).
- Given that \(y = \left( x^2 + \dfrac{2}{x} \right)^8\), find the value of \(\dfrac{dy}{dx}\) when \(x = -1\).
By the
Chain Rule we have that
\[\begin{align*}
\frac{dy}{dx} &= 8 \left( x^2 + \frac{2}{x} \right)^7 \times \frac{d}{dx} \left( x^2 + \frac{2}{x} \right) \\
&= 8 \left( x^2 + \frac{2}{x} \right)^7 \left( 2x - 2x^{-2} \right).
\end{align*}\]
So we have
\[\begin{align*}
\frac{dy}{dx}(-1) &= 8 \left( (-1)^2 + \frac{2}{-1} \right)^7 \left( 2(-1) - 2(-1)^{-2} \right) \\
&= 8 (1-2)^7 (-2-2) \\
&= 8 (-1)^7 (-4) \\
&= 32.
\end{align*}\]
