Precalculus 4 - Theorem 1.2 proof, possible correction

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Precalculus 4 - Theorem 1.2 proof, possible correction

msteiner
For the proof of Theorem 1.2, toward the end of the proof to show symmetry about the line x = h, the text states:

To show that the graph is symmetric about the line x = h, we need to show that if we move left or right the same distance away from x = h, then we get the same y-value on the graph. Suppose we move ∆x to the right or left of h. The y-values are the function values so we need to show that F(a + ∆x) = F(a − ∆x).

Shouldn't that be that we need to show that F(h + ∆x) = F(h − ∆x)? And shouldn't that follow in the rest of the proof as follows?

Given that
F(h+∆x) = a|h+∆x−h|+k = a|∆x|+k
and
F(h−∆x) = a|h−∆x −h|+k = a|−∆x|+k = a|∆x|+k

we see that F(h + ∆x) = F(h − ∆x). Thus we have shown that the y-values on the graph on either side of
x = h are equal provided we move the same distance away from x = h.

I hope I'm not just overlooking something here. Thanks.
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Re: Precalculus 4 - Theorem 1.2 proof, possible correction

Carl Stitz
Administrator
Yes!

Thank you so very much!

Carl



From: msteiner [via Stitz Zeager Open Source Mathematics] <[hidden email]>
To: Carl Stitz <[hidden email]>
Sent: Monday, October 15, 2018 11:53 PM
Subject: Precalculus 4 - Theorem 1.2 proof, possible correction

For the proof of Theorem 1.2, toward the end of the proof to show symmetry about the line x = h, the text states:

To show that the graph is symmetric about the line x = h, we need to show that if we move left or right the same distance away from x = h, then we get the same y-value on the graph. Suppose we move ∆x to the right or left of h. The y-values are the function values so we need to show that F(a + ∆x) = F(a − ∆x).

Shouldn't that be that we need to show that F(h + ∆x) = F(h − ∆x)? And shouldn't that follow in the rest of the proof as follows?

Given that
F(h+∆x) = a|h+∆x−h|+k = a|∆x|+k
and
F(h−∆x) = a|h−∆x −h|+k = a|−∆x|+k = a|∆x|+k

we see that F(h + ∆x) = F(h − ∆x). Thus we have shown that the y-values on the graph on either side of
x = h are equal provided we move the same distance away from x = h.

I hope I'm not just overlooking something here. Thanks.



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