Can we design a polynomial in \(x\), \(p(x)\), such that \(p(x)=p'(x)\)?

To tackle this question we could look at the prompting questions.

- What is the simplest polynomial in \(x\) you can think of and what is its derivative?

The simplest polynomial in \(x\) could be \(p(x)=\text{constant}\). If this were the case then \(p'(x)=0\), so if the constant equals zero then we would have a solution. This solution does seem a silly one, can we find another?

We could choose \(p(x)=c+x\), where \(c\) is a constant. What would \(c\) need to be to make \(p'(x)\) have the same constant term as \(p(x)\) in this case?

- What powers of \(x\) do you want the polynomial to include and what does this mean for the derivative?

If we choose to include an \(x^2\) term in our polynomial \(p(x)\), then \(p'(x)\) would include an \(x\) term. From part (a), if we choose \(p(x)=1+x\) then we have \(p(x)\) agreeing with \(p'(x)\) as far as the \(x^0\) term. So if we include \(\dfrac{x^2}{2}\) in \(p(x)\), we would then get \[p(x)=1+x+\frac{x^2}{2}\] and \[p'(x)=1+x,\] so that \(p(x)\) and \(p'(x)\) agree as far as the \(x^1\) term (but not beyond).

How could we continue this?

What do the polynomials you find look like when plotted, \(y=p(x)\)?

You may want to use Desmos to plot the function \(p(x)\) as you construct it.