|
Example 1
Which function corresponds to the graph shown below?
-
$y=\dfrac{1}{2}\cos(x)$
-
$y=-\dfrac{1}{2}\cos(x)$
-
$y=\dfrac{1}{2}\sin(4x)$
-
$y=-\dfrac{1}{2}\sin(4x+\pi/2)$
-
$y=-\dfrac{1}{2}\cos(4x+\pi)$
Solution
- Let us start by evaluating at $x=0$ the 5 functions given as possible answers.
-
$y(0)=\dfrac{1}{2}\cos(0)=\dfrac{1}{2}$
-
$y(0)=-\dfrac{1}{2}cos(0)=-\dfrac{1}{2}$
-
$y(0)=\dfrac{1}{2}\sin(4(0))=\dfrac{1}{2}\sin(0)=0$
-
$y(0)=-\dfrac{1}{2}\sin(4(0)+\pi/2)=-\dfrac{1}{2}\sin(0+\pi/2)=-\dfrac{1}{2}\sin(\pi/2)=-\dfrac{1}{2}$
-
$y=-\dfrac{1}{2}cos(4(0)+\pi)=-\dfrac{1}{2}\cos(4(0)+\pi)=-\dfrac{1}{2}\cos \pi=-\dfrac{1}{2}(-1)=\dfrac{1}{2}$
- It can be seen that the answer can only be $A$ or $E$. We now compare the periods of function $y=\dfrac{1}{2}\cos(x)$ in $A$ which is $2\pi$ and the period of function $y=-\dfrac{1}{2}\cos(4x+\pi)$ in E which is $\dfrac{2\pi}{4}=\dfrac{\pi}{2}$ which corresponds to the period of the graph.
Answer E
|