Can we sketch the graph of this piecewise function?

Can we sketch the graph of this piecewise function?

Question

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 $x=0$, $1$, $2$.

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1. For $1 \le t \le 2$, define $g(t) = \int_{t-1}^{t} f(x) \ dx.$ Express $g(t)$ as a cubic in $t$.

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1. Calculate and factorise $g'(t)$.

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1. What are the minimum and maximum values of $g(t)$ for $t$ in the range $1 \le t \le 2$?

Suggestion

Are there any turning points for $f(x)$?

1. For $1 \le t \le 2$, define $g(t) = \int_{t-1}^{t} f(x) \:dx.$ Express $g(t)$ as a cubic in $t$.

Can we helpfully split $g(t)$ into two integrals?

What happens if we differentiate $g(t)$?

Where could $g(t)$ attain its minimum and maximum values?

Solution

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

We have

$$\begin{align*} g'(t) &= 2t^2 - 7t + 6 \\ &= (2t-3)(t-2). \end{align*}$$

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 $\dfrac{3}{2}$.