<p>The curve $y=\mathrm{f}\left( x \right)$ for $0\le x<3$ is shown below. It is given that $\mathrm{f}$ is an odd function, $\mathrm{f}\left( -x \right)=-\mathrm{f}\left( x \right)$, and $\mathrm{f}\left( x+6 \right)=\mathrm{f}\left( x \right)$ for all real values of $x$. Sketch the curve of $y=\mathrm{f}\left( x \right)$ for $-3<x\le 6$.</p><p style="text-align: right;">[4]</p>
<figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/converted/1759052558917-69mdilb66ea.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NTkwNTI1NTg5MTctNjltZGlsYjY2ZWEucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzQ5ODQ3NiwiZXhwIjoxNzkzNDk4Nzc2fQ.qMITfeWuvdjD6sjh6twN9Lh0NXDOngMLCPKqZC_G6Bs" width="400" class="aligncenter" /></figure>

The curve $y=\mathrm{f}\left( x \right)$ for $0\le x<3$ is shown below. It is given that $\mathrm{f}$ is an odd function, $\mathrm{f}\left( -x \right)=-\mathrm{f}\left( x \right)$, and $\mathrm{f}\left( x+6 \right)=\mathrm{f}\left( x \right)$ for all real values of $x$. Sketch the curve of $y=\mathrm{f}\left( x \right)$ for $-3<x\le 6$.

Default solution

The graph is a piecewise linear function passing through the following points:

$(-3, -2)$, $(0, 0)$, $(2, 2)$, $(4, -2)$, $(6, 0)$

From $x = -3$ to $x = 2$: Line with gradient $\frac{2-(-2)}{2-(-3)} = \frac{4}{5}$

From $x = 2$ to $x = 4$: Line with gradient $\frac{-2-2}{4-2} = -2$

From $x = 4$ to $x = 6$: Line with gradient $\frac{0-(-2)}{6-4} = 1$

Primary Solution

Periodic Functions

Functions

International Baccalaureate Mathematics: Analysis and Approaches Higher Level

NYJC Basic Graphing Techniques Assignment Q2

NYJC Basic Graphing Techniques Assignment Q2

Video Solution:

Default solution