# 2003_Question4 (i)

Note that f'(x) = 2x - 2p = 2(x - p) so f'(x) = 0 iff x = p. So there is a stationary point in 0 < x < 1 iff 0 < p < 1.

(ii)

If p \ge 1, then f'(x) < 0 for 0 \le x < 1. So f is decreasing on 0 \le x \le 1. So the minimum value attained is at x = 1, ie. is f(1) = 1 - 2p + 3 = 4 - 2p.

(iii)

If p \le 0, then f'(x) \ge 0. So f is increasing on 0 \le x \le 1. So the maximum value is attained at x = 0, so m = 3. Alternatively note that if p \le 0, -2p \ge 0, so f(x) = x^2 - 2px + 3 \ge 3 with equality at x = 0.

(iv)

Note that in general that because f'(x) = 2(x - p), we have that f is decreasing for x < p and increasing for x > p. So, a minimum is attained at x = p. (alternatively release that the second derivative is negative at x = p) So the minimum value in this case is f(p) = p^2 - 2p^2 + 3 = 3 - p^2.

(v)

We know that:

\displaystyle m(p) = \begin{cases}3 & -2 \le p \le 0 \\ 3 - p^2 & 0 < p < 1 \\ 4 - 2p & p \ge 1\end{cases}

for -2 \le p \le 2.

We note that 3 = 3 - 0^2 and 3 - 1^2 = 4 - 2 = 2, so m is certainly continuous in p meaning that we will not have any “jumps”. We expect a straight horizontal line from -2 to 0, into a decreasing quadratic, then finally into a decreasing straight line. This will look like: 