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Worked solution

Repeat the defining interval without losing endpoints

It is given that f(x)={3+x0<x≤2,7−x2<x≤4,f(x)=\begin{cases}3+x&0<x\le2,\\7-x&2<x\le4,\end{cases} and f(x)=f(x+4)f(x)=f(x+4) for all real xx. (i) Find f(1)f(1), f(8)f(8) and f(29)f(29). (ii) Sketch y=f(x)y=f(x) for −3<x≤8-3<x\le8.

CJC · Functions · Example 12

(i) Find the three values

Answer

f(1)=4f(1)=4, f(8)=3f(8)=3, f(29)=4f(29)=4.

f(1)=3+1=4f(1)=3+1=4. Reduce 8 to 4, which belongs to the defining interval (0,4](0,4]: f(8)=f(8−4)=f(4)=7−4=3f(8)=f(8-4)=f(4)=7-4=3. Reduce 29 by 28 to 1: f(29)=f(1)=4f(29)=f(1)=4. Multiples of 4 reduce to 4, not to an undefined base input 0.

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(ii) Sketch on the requested interval

Answer

Join (−3,4)(-3,4) open, (−2,5)(-2,5), (0,3)(0,3), (2,5)(2,5), (4,3)(4,3), (6,5)(6,5) and (8,3)(8,3) closed by straight segments.

Translate the defining two line segments by integer multiples of 4 and clip to −3<x≤8-3<x\le8. The first endpoint is excluded by the requested sketch interval; the last is included. All interior peaks and troughs are included. Although the base rule excludes 0, the periodic extension defines f(0)=f(4)=3f(0)=f(4)=3. Follow the source remark: label endpoint coordinates and each point where the segments meet.

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Period 4 triangular graph on −3<x≤8: open (−3,4), peaks (−2,5),(2,5),(6,5), troughs (0,3),(4,3), closed (8,3); all interior joins included.
Period 4 triangular graph on −3<x≤8: open (−3,4), peaks (−2,5),(2,5),(6,5), troughs (0,3),(4,3), closed (8,3); all interior joins included.
Period 4 triangular graph on −3<x≤8: open (−3,4), peaks (−2,5),(2,5),(6,5), troughs (0,3),(4,3), closed (8,3); all interior joins included.
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