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 \(m=4-2p\).

  2. What is the value of \(m\) if \(p \le 0\)?

  3. Obtain a formula for \(m\) in terms of \(p\), valid for \(0 < p < 1\).

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

Set of axes with p going from minus 2 to 2 and with m going from - 1 to 3, with m being the vertical axis.