Porisma

Worked solution

Reduce periodic inputs and sketch a complete finite window

It is given that

f(x)={7−x2for 0<x≤2,2x−1for 2<x≤4f(x)=\begin{cases}7-x^2&\text{for }0<x\leq2,\\2x-1&\text{for }2<x\leq4\end{cases}

and that f(x)=f(x+4)f(x)=f(x+4) for all real values of xx.

(i) Evaluate f(27)+f(45)f(27)+f(45). marks

(ii) Sketch the graph of y=f(x)y=f(x) for −7≤x≤10-7\leq x\leq10. marks

[N2009/I/4 (i), (ii)]

CJC · Review Problems · Question 7

Answer

(i) 1111. (ii) Repeat the quadratic/linear shape with period 4 over [−7,10][-7,10]; peaks 7, troughs 3 and left endpoint (−7,6)(-7,6).

(i)

Repeatedly subtract 4: f(27)+f(45)=f(23)+f(41)=f(19)+f(37)=⋯=f(3)+f(1)f(27)+f(45)=f(23)+f(41)=f(19)+f(37)=\cdots=f(3)+f(1). The linear branch gives f(3)=5f(3)=5 and the quadratic branch gives f(1)=6f(1)=6, so the sum is 1111.

How did this part go?

(ii)

Repeat the base graph every 4 units. On each 4k<x≤4k+24k<x\leq4k+2, draw y=7−(x−4k)2y=7-(x-4k)^2; on each 4k+2<x≤4k+44k+2<x\leq4k+4, draw y=2(x−4k)−1y=2(x-4k)-1. Restrict the drawing to [−7,10][-7,10]. The peaks are (−4,7),(0,7),(4,7),(8,7)(-4,7),(0,7),(4,7),(8,7), the troughs are (−6,3),(−2,3),(2,3),(6,3),(10,3)(-6,3),(-2,3),(2,3),(6,3),(10,3), and the left endpoint is (−7,6)(-7,6). The joins are continuous, and both window endpoints are included.

How did this part go?

Compare your sketch

Expand diagram

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Period-four graph on −7≤x≤10: starts at (−7,6), alternates curved descents from height 7 to 3 and straight ascents from 3 to 7. Peaks at x=−4,0,4,8; troughs at x=−6,−2,2,6,10. All joins and the two window endpoints are included.
Period-four graph on −7≤x≤10: starts at (−7,6), alternates curved descents from height 7 to 3 and straight ascents from 3 to 7. Peaks at x=−4,0,4,8; troughs at x=−6,−2,2,6,10. All joins and the two window endpoints are included.
Period-four graph on −7≤x≤10: starts at (−7,6), alternates curved descents from height 7 to 3 and straight ascents from 3 to 7. Peaks at x=−4,0,4,8; troughs at x=−6,−2,2,6,10. All joins and the two window endpoints are included.

Common mistakes

A multiple of 4 reduces to 4, not to the excluded base input 0; periodicity supplies the value 7 at each join.

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