Question

Show that the local maximum value of $x+\frac{1}{x}$ is less than the local minimum value.

Answer 1

Step 1. Determine the value of $x$.

Consider the given function $y=x+\frac{1}{x}$.

The first derivative is $\frac{dy}{dx}=1-\frac{1}{x^2}$.

Set $\frac{dy}{dx}=0$ to determine the value of $x$.

$\begin{array}{rcl}1-\frac{1}{x^2}& =& 0\\ & \Rightarrow & \frac{1}{x^2}=1\\ & \Rightarrow & x^2=1\\ & \Rightarrow & x=\pm 1\end{array}$

The value is $x=\pm 1$

Step 2. Prove that the local maximum value is less than the local minimum value.

The second derivative is $\frac{d^{2}y}{d{x}^2}=\frac{2}{x^3}$.

The function can have either max or min values at $1\mathrm{and}-1$.

The second derivative test will help identifying if it is max or min at each value.

The second derivative is $\frac{d^{2}y}{d{x}^2}=\frac{2}{x^3}$.

There are two cases:

1. If $f\text{'}\text{'}\left(x\right)<0$, the function is maximum at $x=a$.
2. If $f\text{'}\text{'}\left(x\right)>0$, the function is minimum at $x=a$.

So, substitute $x=1$ into $\frac{d^{2}y}{d{x}^2}$.

$\begin{array}{rcl}\frac{d^{2}y}{d{x}^2}& =& \frac{2}{{\left(1\right)}^3}\\ & \Rightarrow & \frac{d^{2}y}{d{x}^2}=2\\ & \Rightarrow & \frac{d^{2}y}{d{x}^2}>0\end{array}$

Now, substitute $x=-1$ into $\frac{d^{2}y}{d{x}^2}$.

$\begin{array}{rcl}\frac{d^{2}y}{d{x}^2}& =& \frac{2}{{\left(-1\right)}^3}\\ & \Rightarrow & \frac{d^{2}y}{d{x}^2}=-2\\ & \Rightarrow & \frac{d^{2}y}{d{x}^2}<0\end{array}$

Therefore, the local maximum value of $y$ is at $x=-1$ and the local maximum value is $-2$ and the local minimum value of $y$ is at $x=1$ and the local minimum value is $2$.

Hence, the local maximum value $\left(-2\right)$ is less than the local minimum value $\left(2\right)$.