If $f(x)=x^3-3x^2-x+2$ ,Find an expression for the function $y$, which is obtained by rotating the graph of $f (x)$ through $180°$.

Question

Question: $f(x)=x^3-3x^2-x+2$ where $x\leq 1$

Find an expression for the function $y = g(x)$, where $x\geq1$ , which is obtained by rotating the graph of $y = f (x)$ through $180°$ about the point $(1, –1)$.

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So for this question I don't really understand the wording , what does it mean by rotating the graph $y=f(x)$ through $180°$ about the point $(1,-1)$?

The only progress I made on this question was to find a point of inflection of $f(x)$ which is

$$f'(x)=3x^2-6x-1 \Leftrightarrow f''(x)=6x-6=0 \Leftrightarrow ~\therefore x=1$$

But I am not sure what to do with that...

Answer 1

The graph of the function is shown below:

It is clear that if we rotate the part $\{(x,f(x)):x \geq 1\}$ we get the portion of the graph $\{(x,f(x)): x \leq 1\}$. Hence $g = f$. This can also be seen as follows: If we rotate the point $P(x,f(x))$ on the curve by $180^\circ$ about $(1,-1)$, then we obtain the point $(2-x, -2-f(x))$. Note that $f(2-x)+f(x) = -2$ and hence $f(2-x) = - 2 - f(x)$ and $g= f$.

Answer 2

Here's a general answer: rotation of the graph of a function $f$ around a point $(a,b)$ is but the symmetry of the graph w.r.t. this point, hence it is $$y=g(x)\stackrel{\text{def}}{=}2b-f(2a-x).$$

To see this, denote $x,y=f(x)$ a point on the original graph, and $(X,Y=g(X)$ athe symmetric point. By definition, we have $X=2a-x$, $Y=2b-y$ (writing the point $(a,b)$ is the middle point of $(x,y)$ and $(X,Y)$).