Designing a polynomial

Designing a polynomial

Problem

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

Prompting questions

1. What is the simplest polynomial in $x$ you can think of and what is its derivative?

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1. What powers of $x$ do you want the polynomial to include and what does this mean for the derivative?

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

Things you might have tried

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

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?

Answer

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).

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.