Review question

# Can we find the minimum value of $x^2+2/x$ for $x$ positive? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5640

## Solution

1. 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 $448$.

1. 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 $1^2+\dfrac{2}{1} = 3$.

1. 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*}