Let $f(x) = x^3-3x^2+6$ for $x\in \mathbb{R}$ and $g(x) = \max\{f(t); x+1\leq t\leq x+2\}$ for $-3\leq x\leq0$. Need help in understanding $g(x)$.

Question

I came across this question in my book:

Let $f(x) = x^3-3x^2+6$ for $x\in {R}$ and $g(x) = \max\{f(t); x+1\leq t\leq x+2\}$ , for $-3\leq x \le0$ and $1-x, x\geq 0$. Test continuity of $g(x)$ for $x\in [-3,1]$.

My Problem

I have not been able to attempt the problem since I am stuck on the first step: Identifying the function. I have absolutely no idea how $g(x)$ looks like for $x\in [-3,0]$. I need help in understanding this part: $$g(x)=\max\{f(t); x+1\leq t\leq x+2\}$$

Thanks for comments/suggestions.

Answer 1

Firstly calculate the critical points of $f(x)$.

You get that $f(x)$ has a local maximum on $x=0$ and local minimum on $x=2$ implying it is strictly decreasing in $(0,2)$ and strictly increasing in $(-\infty,0)\cup(2,\infty).$

Now look at the definition of $g(x)$. It says that $g(x)=$ maximum value of $f(x)$ in the interval $[x+1,x+2]$ where $x\in[-3,1]$

Now let's split up the intervals.

Interval I : $x\in[-3,-2]$ then $t\in[-2,0]$

You can see that $f(x)$ is strictly increasing in $[-2,0)$ so it will always have maximum value in $[x+1,x+2]$ at $t=x+2$

$$g(x)=f(x+2)\;\;\;\forall\;\;\;x\in[-3,-2]$$

Interval II : $x\in[-2,-1]$ then $t=0$ will always exist in $[x+1,x+2]$ and as $f(0)$ is the local maxima.

$$g(x)=f(0)=6\;\;\;\forall\;\;\;x\in[-2,-1]$$

Interval III : $x\in[-1,0]$ then $t\in[0,2]$

You can see that $f(x)$ is strictly decreasing in $(2,0)$ so it will always have maximum value in $[x+1,x+2]$ at $t=x+1$

$$g(x)=f(x+1)\;\;\;\forall\;\;\;x\in[-2,0]$$

Interval IV is same as Interval I so you can incorporate this part into that.

I don't how to typeset cases or I would have provided a complete neat answer.

Answer 2

Fix $x \in [-3,0]$ and consider the intervall $I_x:=[x+1,x+2]$. Then consider the set $f(I_x)$. Since $I_x$ is compact and $f$ is continuous, the set $f(I_x)$ has a maximum.

We have $g(x):= \max f(I_x)$

Answer 3

Hint. The derivative of $f$ is $f'(x) = 3x^2-6x=3x(x-2)$, hence $f$ is increasing in $(-\infty,0]$, it is decreasing in $[0,2]$, and again increasing in $[2,+\infty)$. Now you should be able to find $$\max_{t\in[x+1,x+2]} f(t)$$ for any $x\in\mathbb{R}$.