Review question

# Can we sketch the graph of this piecewise function? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7196

## Solution

Let $\quad f(x) = \begin{cases}x+1 & \text{for } 0 \leq x \le 1; \\ 2x^2-6x+6 & \text{for } 1 \leq x \le 2. \end{cases}$

1. Sketch a graph of $y=f(x)$ for $0 \le x \le 2$, labelling any turning points and the values attained at $x=0$, $1$, $2$.

There is no turning point in the interval $0 \le x \le 1$. To find any turning points in the interval $1 \le x \le 2$, we first differentiate $f(x)$ in this interval to get $f'(x) = 4x - 6.$ Setting $f'(x)=0$ gives $x=\dfrac{3}{2}$, so there is just one turning point, at $\left(\dfrac{3}{2},\dfrac{3}{2}\right)$.

We also have $f(0) = 1; \quad f(1) = 2; \quad f(2) = 2.$

We could alternatively find the turning point for the quadratic section by completing the square. We have $2x^2-6x+6= 2(x - 3/2)^2 + 3/2$.

1. For $1 \le t \le 2$, define $g(t) = \int_{t-1}^{t} f(x) \:dx.$ Express $g(t)$ as a cubic in $t$.
As $0 \le t-1 \le 1 \leq t$, we can split the integral at $x=1$. We have \begin{align*} g(t) &= \int_{t-1}^{1} x+1 \:dx + \int_{1}^{t} 2x^2 - 6x + 6 \:dx \\ &= \left[\dfrac{x^2}{2}+x\right]_{t-1}^{1} + \left[\dfrac{2x^3}{3} - 3x^2 + 6x \right]_{1}^{t} \\ &= \dfrac{2t^3}{3} - \dfrac{7t^2}{2} + 6t - \dfrac{5}{3}. \end{align*}

Note $\displaystyle\int_{t-1}^{1} x+1 \:dx = \left[\dfrac{x^2}{2}+x\right]_{t-1}^{1}= \left[\dfrac{(x+1)^2}{2}\right]_{t-1}^{1}$.

1. Calculate and factorise $g'(t)$.
We have \begin{align*} g'(t) &= 2t^2 - 7t + 6 \\ &= (2t-3)(t-2). \end{align*}
1. What are the minimum and maximum values of $g(t)$ for $t$ in the range $1 \le t \le 2$?

The minimum and maximum values will occur either at the end points ($t=1$ or $t=2$) or at the turning points.

The turning points occur when $g'(t)=0$, which is when $t=\dfrac{3}{2}$ or $t=2$.

We have \begin{align*} g(1) &= \dfrac{3}{2} = \dfrac{36}{24}; \\ g\left(\dfrac{3}{2}\right) &= \dfrac{41}{24}; \\ g(2) &= \dfrac{5}{3} = \dfrac{40}{24}. \end{align*}

So the maximum occurs at $t=\dfrac{3}{2}$ with value $\dfrac{41}{24}$, and the minimum is at $t=1$ with value $\dfrac{3}{2}$.