Things you might have tried

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.

  1. 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?

If we could choose \(p(x)=c+x\), where \(c\) is a constant then this would give \(p'(x)=1\). Thus, if we choose \(c\) to be \(1\) we would have \(p(x)\) and \(p'(x)\) agreeing as far as the \(x^0\) (constant) term (but they disagree beyond that).

  1. 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.