Question

What is the minimum value of `g(x) = x/csc(pi*x)` on the interval `[0,1]`?

Answer 1

There is a minimum value of `0` located both at `x=0` and `x=1`.

Explanation:

First, we can immediately write this function as

`g(x)=x/(1/sin(pix))=xsin(pix)`

Recalling that `csc(x)=1/sin(x)`.

Now, to find minimum values on an interval, recognize that they could occur either at the endpoints of the interval or at any critical values that occur within the interval.

To find the critical values within the interval, set the derivative of the function equal to `0`.

And, to differentiate the function, we will have to use the product rule. Application of the product rule gives us

`g'(x)=sin(pix)d/dx(x)+xd/dx(sin(pix))`

Each of these derivatives give:

`d/dx(x)=1`

And, through the chain rule:

`d/dx(sin(pix))=cos(pix)*underbrace(d/dx(pix))_(=pi)=picos(pix)`

Combining these, we see that

`g'(x)=sin(pix)+pixcos(pix)`

Thus, critical values will occur whenever

`sin(pix)+pixcos(pix)=0`

We can't solve this algebraically, so use a calculator to find all of this function's zeros on the given interval `[0,1]`:

graph{sin(pix)+pixcos(pix) [-.1, 1.1, -3, 2.02]}

The two critical values within the interval are at `x=0` and `xapprox0.6485`.

So, we know that the minimum value of `g(x)` could occur at `3` different places:

• `x=0` or `x=1`, the endpoints of the interval
• `x=0` or `x=0.6485`, the critical values within the interval

Now, plug in each of these possible values into the interval:

`{(g(0)=0,color(red)text(minimum)),(g(0.6485)=0.5792,color(blue)text(maximum)),(g(1)=0,color(red)text(minimum)):}`

Since there are two values which are equally low, there are minima both at `x=0` and `x=1`. Note that even though we went through of finding the trouble of `x=0.6485`, it wasn't even a minimum.

Graphed is `g(x)` on the interval `[0,1]`:

graph{x/csc(pix) [-.05, 1.01, -.1, .7]}

Also, note that the minimum value is `0`, since `g(0)=g(1)=0`. The distinction is that `x=0` and `x=1` are the locations of the minima.