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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?

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  • $\begingroup$ Essentially you used the average of the $x$ intercepts of the parabola to find $x=3$.You can then check and see that $f(3+h)=f(3-h)$ for all $h$ $\endgroup$ – imranfat May 8 '18 at 21:53
  • $\begingroup$ Check this: a parabola of this form has two zeros, i.e. $x = 0$ and $x = 6$. Since it is symmetric, the maximal value corresponds to $x = 3$ - the middle point of the zeros. $\endgroup$ – Gibbs May 8 '18 at 21:55
  • $\begingroup$ The only solution of $x = 6 - x$ is? $\endgroup$ – NickD May 8 '18 at 21:58
  • $\begingroup$ This can be made to work once you add/prove the necessary assumptions about concave parabolas. But you have to be careful in general, try for example to apply the same argument to the very similarly looking $\,f(x)=x^2-6x\,$, or $\,f(x)=x^2(6-x)^2\,$, and see why/where it fails. $\endgroup$ – dxiv May 8 '18 at 22:09
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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 $6-x_0=x_0$.

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  • $\begingroup$ This is a good answer that actually addresses some of the symmetry. (+1) $\endgroup$ – T. Bongers May 8 '18 at 21:59
  • $\begingroup$ Thank you, this is a clear answer. Do you know if the same argument works a function that has higher orders? For example $f(x)=50x^2-x^4$ $\endgroup$ – matryoshka May 8 '18 at 22:11
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    $\begingroup$ @Grace I'm not sure about this. With a first look, i think this one has 2 maxima. It is $f(x)=x^2(50-x^2)=f(50-x^2)$, so I believe the maxima are located at the solutions of $x=50-x^2$. I'm not sure about this though. $\endgroup$ – JustDroppedIn May 8 '18 at 22:14
  • $\begingroup$ Let consider $f(x)=50x^2-x^4=625-625+50x^2-x^4=625-(25-x^2)^2$ $\endgroup$ – gimusi May 9 '18 at 5:41
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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 $x=3$.

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    $\begingroup$ Squared, not cubed. $\endgroup$ – NickD May 8 '18 at 21:57
  • $\begingroup$ opssss...typo! I fix. Thanks $\endgroup$ – gimusi May 8 '18 at 21:57
  • $\begingroup$ I've added some trivial details on symmetry for the less perceptive downvoters. $\endgroup$ – gimusi May 8 '18 at 22:05
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HINT: Because it is a polynomial of second degree, its extremum lies on the symmetry line of the interval with ends at its zeroes.

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