Where does $f(x) = x^2 - 2px + 3$ have its minimum?

Where does $f(x) = x^2 - 2px + 3$ have its minimum?

Question

In this question we shall consider the function $f(x)$ defined by $f(x) = x^2 - 2px + 3$ where $p$ is a constant.

1. Show that the function $f(x)$ has one stationary value in the range $0< x <1$ if $0< p <1$, and no stationary values in that range otherwise.

In the remainder of the question we shall be interested in the smallest value attained by $f(x)$ in the range $0 \le x \le 1$. Of course, this value, which we shall call $m$, will depend on $p$.

1. Show that if $p \ge 1$ then $m=4-2p$.

---

1. What is the value of $m$ if $p \le 0$?

---

1. Obtain a formula for $m$ in terms of $p$, valid for $0 < p < 1$.

---

1. Using the axes opposite, sketch the graph of $m$ as a function of $p$ in the range $-2 \le p \le 2$.

Solution

We know that $y = f(x)$ is a parabola, and since the coefficient of $x^2$ is positive, it is ‘vertex-down’ and has a minimum. It can only have one stationary value.

Differentiating $f(x)$ gives $f'(x) = 2x - 2p,$ so the stationary point occurs when $x=p$, which is in the range $0<x<1$ if and only if $0<p<1$.

Or else completing the square, $f(x) = (x-p)^2 + 3 - p^2$, and so the vertex of the parabola is at $(p, 3-p^2)$, which is in the range $0<x<1$ if and only if $0<p<1$.

The minimum value $m(p)$ attained by $f(x)$ in the range $0 \le x \le 1$ will occur either at an endpoint ($x=0$ or $x=1$), or at a stationary point in between 0 and 1.

If $p \ge 1$ then we have seen there is no stationary point in that interval. So as $f(1)=4-2p<3=f(0)$, we have $m(p)=4-2p$.

For $p \le 0$, again, no stationary point exists in the range $0 \le x \le 1$. We now have $f(0) = 3 < 4 - 2p = f(1)$, so $m(p)=3$.

If $0 < p < 1$, then we know there is a stationary point of $f$ at $x=p$, which is at the parabola’s vertex.

So the minimum value $m(p)=f(p)=3-p^2.$

To summarise the results of parts ii), iii) and iv) the function $m(p)$ is: $m(p)= \begin{cases} 3 &\text{if } p \le 0 \\ 3-p^2 &\text{if } 0 < p < 1 \\ 4-2p &\text{if } 1 \le p \end{cases}$

A sketch of this looks like

How do we know for sure that the join at $p=1$ is smooth? Because the gradient of $3-p^2$ is $-2p$ and the gradient of $4-2p$ is $-2,$ which agree when $p=1$.

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/h6xZbc2V/width/600/height/400/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

Download GeoGebra file