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F(x)=1 if the decimal expansion of x contains a 5; =0 otherwise.

According to the answer, x is a local minimum point if its decimal expansion does not contain 5.

If I understand the question correctly, if x’s decimal expansion does not contain 5, x should be 0 and every number around it should also be 0 unless x is .49999999 So the correct answer should be either local minimum or both a local minimum and a local maximum

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  • F(x) is a piecewise function assuming only two values: F(x)=1 when the decimal expansion of x contains a 5, and F(x)=0 otherwise. Since F(x) does not assume any other value, every x which does not have 5 in its decimal expansion serves as a local minimum, and every x which does serves as a local maximum.
    – Manan
    Commented Jul 7, 2020 at 7:24
  • There is a third option. Every x is both a maximum and a minimum no?
    – jxhyc
    Commented Jul 7, 2020 at 7:26
  • I don't think a value from the domain can simultaneously be a local minimum and maximum. Why would you think it can be both?
    – Manan
    Commented Jul 7, 2020 at 7:28
  • For the situation where x is a horizontal line
    – jxhyc
    Commented Jul 7, 2020 at 7:29
  • From page 635 answer of calculus:Michael Slovak edition: in all other cases, x is both a local maximum and a local minimum
    – jxhyc
    Commented Jul 7, 2020 at 7:31

1 Answer 1

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As elaborated by you in the comments, I looked up for the exact question and solution provided in Spivak's Calculus, 3rd Edition. Perhaps the way the solution was phrased(for the sake of compactness) confused you. Here's the solution:

x is a local maximum(minimum) point if the decimal expansion contains(does not contain) a 5.

This simply means that for every x containing a 5 in its decimal expansion, the function acquires a local maximum(=1), and for every x not containing a 5, the function acquires a local minimum(=0).

Furthermore, a point cannot serve as a local minimum and maximum simultaneously. I suggest you to go through the Calculus 1 course on Khan Academy.

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  • Very strange. I have the 4th edution and there seesm to be some change.
    – jxhyc
    Commented Jul 8, 2020 at 12:39
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    @jxhyc I find the solution in 4th edition to be quite baffling. I've always seen Spivak as the gold-standard Calculus textbook so I wouldn't expect an error like this one. I'll pass and upvote the question for now. Maybe an expert can enlighten me as well.
    – Manan
    Commented Jul 8, 2020 at 16:33

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