E.g. to find the extremum of f(x)=x−x2 we can notice that f(x)=6x−x2=x(6−x)=(6−x)x. This operation maps every value x to 6−x through the axis of symmetry and vice-versa (e.g. 0 is mapped to 6 and 6 mapped to 0). This operation preserves symmetry for x=3 the axis of symmetry and hence the extremum of the function.What about the function f(x)=6x+x2=x(6+x)=(6+x)x? We know that the minimum of this function is at x=−3 but 6+(−3)=3. Also this operation maps e.g 4 to 10, but 10 to 16! Hence, this approach fails with this example (symmetry is not preserved).Why is that so? Why is this approach not working for all the quadratic functions?

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Drake Huang · Accepted Answer

It will work, but you need to find the correct symmetry transformation. Let&#039;s talk more about the first example, f(x)=x(6−x). The symmetry transformation here is x→6−x because if we replace every x in x(6−x) with 6−x we get(6−x)(6−(6−x))=(6−x)(6−6+x)=(6−x)x=x(6−x)=f(x).In other words, f(6−x)=f(x). We do the transformation, in this case replace x with 6−x, and arrive at the original function.Now for your second example, f(x)=x(6+x). To use the same argument we need to find the right transformation. In this case, the transformation is x↦−x−6. We can verify this:f(−x−6)=(−x−6)(6−x−6)=(−x−6)(−x)=(x+6)x=x(6+x)=f(x).Furthermore, the transformation x↦−x−6 maps 4 to -10 and -10 to 4, and we can verify symmetry in general as well. In this case the only value that doesn&#039;t change is x=−3, because −(−3)−6=−3. So, the axis of symmetry, as well as the minimum of the function, is at x=−3.

Matronola3zw6 · Accepted Answer

4 doesn&#039;t get mapped to 10. 4 gets mapped to -10. And -10 gets mapped to 4. It works.The axis of symmetry is x=−3. So if you have a point s=−3+M it will be mapped to −3−M.So 4=−3+M so M=7. So 4↦−3−7=−10. And −10=−3+M′ so M′=−7 so −10→−3−(−7)=4.More generally if x=−3+M then M=x+3 so x↦−3−(x+3)=−6−x.And so 4↦−6−4=−10. And −10↦−6−(−10)=4.

E.g. to find the extremum of \(\displaystyle{f{{\left({x}\right)}}}={x}-{x}^{{{2}}}\)

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