Using symmetry to find the maximum of a function

Question

Without using calculus I was told to find the maximum of $$f(x)=6x-x^2.$$

The "symmetry" approach notices that $f(x)=x(6-x)$, and that replacing $x\leftrightarrow6-x$ does not change $f(x)$ which means it doesn't change the maximum. Then the solution says that the only value of $x$ that is unchanged by $x\leftrightarrow 6-x $ is $x=3$. So that's the location of the maximum.

This is all the solution says. I am under the impression that if $f(x)$ is unchanged then no number should be affected unchanged. How did they figure out that $x=3$ is the only thing unchanged?

Answer 1

Since $f(x)=x(6-x)$, it is $f(6-x)=(6-x)x=f(x)$. Now let $x_0$ be a maximum of $f$. We have $f(x_0)=f(6-x_0)$, therefore since this maximum is attained only one time (e.g. the graph is a parabola), it must be $6-x_0=x_0$.

Answer 2

HINT

Note that

$$f(x)=6x-x^2=9-9+6x-x^2=9-(x-3)^2$$

thus $f(x)$ is the sum of

• a constant term $9$
• a function $-(x-3)^2$ symmetric with respect to $x=3$

and then $f(x)$ itself is symmetric with respect to $x=3$.

Answer 3

HINT: Because it is a polynomial of second degree, its extremum lies on the symmetry line of the interval with ends at its zeroes.