Solution assignment 08 Optimizing f(x)

Assignment 8

Verify whether the function:

$y=\displaystyle\frac{x^2+1}{x}$

has a maximum or a minimum.

Solution

Naturally we have $x\neq0$ .

Candidates for a maximum or a minimum can be found by making the first derivative equal to $0$ and solve the resulting equation. We find the derivative using the quotient rule:

$y'=\displaystyle\frac{x\cdot2x-(x^2+1)\cdot1}{x^2}=\displaystyle\frac{x^2-1}{x^2}=0$

A fraction equals $0$ if its numerator equals $0$ and thus we have to solve:

$x^2-1=0$

and thus:

$x=1$ or $x=-1$

In order to investigate whether a point is a maximum or a minimum we could use the second derivative. However this requires a lot of calculations. It is easier to consider the sign of the first derivative. The denominator $x^2$ is always positive, we only have to look at the numerator. This contains a quadratic function of which the graph is an 'opens up' parabola. Around the left intersection point $x=-1$ the first derivative goes from + via $0$ to - and thus there is a maximum for $x=-1$ . For $x=1$ the reverse holds and thus we have a minimum for this point.
The extremes are thus $(-1,-2)$ (maximum) and $(1,2)$ (minimum).
This result seems strange: a minimum that is greater than a maximum but it is not. Look at the graph below. Notice that the line $y=x$ is an asymptote.