# 2004_Question3

(a)

Note that 1 + 1 = 2 - 6 + 6 = 2, so f is continuous at x = 1. Note that the derivative of y = 2x^2 - 6x + 6 wrt x is 4x - 6, and at x = 0 this is equal to -2 < 0. So f is increasing until x = 1, then decreasing at x = 1. y = 2x^2 - 6x + 6 then has a turning point at x = \dfrac 3 2, after which f is increasing. So we want to draw a straight line from (0,1) to (1,2), leading into a quadratic curve decreasing until x = \dfrac 3 2, and then increasing until x = 2.

We’ve already noted a turning point at x = \dfrac 3 2, (ie. the point \displaystyle \left(\frac 3 2, \frac 3 2\right)) but there is also a turning point at x = 1 (ie. the point (1, 2), even though f is not differentiable there) since f is increasing to the left of 1 and decreasing to the right of 1. We have f(0) = 1, f(1) = 2 and f(2) = 2 \times 4 - 6 \times 2 + 6 = 2.

Since I don’t have paper+pen handy I’ll defer to desmos.

(b)

We have:

\displaystyle g(t) = \int_{1 - t}^1 (x + 1) \mathrm dx + \int_1^t (2x^2 - 6x +6) \mathrm dx

We then have:

\begin{align*}\int_{t - 1}^1 (x + 1) \mathrm dx & = \left[\frac {x^2} 2 + x\right]_{t - 1}^1 \\ & = \frac 3 2 - \frac {(t - 1)^2} 2 - (t - 1) \\ & = \frac 1 2 \left(3 - (t - 1)^2 - 2(t - 1)\right) \\ & = \frac 1 2 \left(3 - (t^2 - 2t + 1) - 2t + 2\right) \\ & = \frac 1 2 \left(-t^2 + 2t - 1 + 3 + 2 - 2t\right) \\ & = -\frac 1 2 t^2 + 2\end{align*}

and:

\begin{align*}\int_1^t (2x^2 - 6x + 6) \mathrm dx & = \left[\frac 2 3 x^3 - 3x^2 + 6x\right]_1^t \\ & = \frac 2 3 t^3 - 3t^2 + 6t - \frac 2 3 + 3 - 6 \\ & = \frac 2 3 t^3 - 3t^2 + 6t - \frac {11} 3\end{align*}

So that:

\displaystyle g(t) = \frac 2 3 t^3 - \frac 7 2 t^2 + 6t - \frac 5 3

We have:

g'(t) = 2t^2 - 7t + 6

Alternatively we could note that by the fundamental theorem of calculus:

g'(t) = f(t) - f(t - 1) = (2t^2 - 6t + 6) - t = 2t^2 - 7t + 6

since if 1 \le t \le 2 then 0 \le t - 1 \le 1.

We can factorise this to:

(2t - 3)(t - 2)

(d)

Note that reading off the derivative of g', g is strictly increasing for t < \dfrac 3 2, has a stationary point at t = \dfrac 3 2, and then decreases to t = 2. Clearly, g must therefore have a maximum at t = \dfrac 3 2, where \displaystyle g(t) = \frac {41} {24}.

Since g is increasing on 1 \le t \le \dfrac 3 2 and decreasing on \dfrac 3 2 \le t \le 2, the minimum on 1 \le t \le \dfrac 3 2 occurs at 1, and is g(1) = \dfrac 3 2, and the minimum on \dfrac 3 2 \le t \le 2 occurs at 2, and is g(2) = \dfrac 5 3. We see \displaystyle \frac 3 2 = \frac {4.5} 3 \le \frac 5 3 so the minimum of g is \dfrac 3 2 (occuring at t = 1), and the maximum is \dfrac {41} {24} (occuring at t = \dfrac 3 2)