Periodic functions > Periodicity
123456Periodicity

## Solutions to the exercises

Exercise 1
a

$f\left(25\right)=f\left(1\right)=5$.

b

$x=2+k\cdot 3$.

c

$x=1\vee x=4\vee x=7\vee x=2,5\vee x=5,5\vee x=8,5$.

Exercise 2
a

$h\left(4,5\right)=-1$, $h\left(10,5\right)=-1$ en $h\left(16,5\right)=-1$.

b

$h\left(0,75\right)=\frac{1}{2}\sqrt{2}$ because you get a rectangular triangle with a ${45}^{o}$ angle that is isosceles with legs $\sqrt{\frac{1}{2}}=\frac{1}{2}\sqrt{2}$.

c

They are all $\frac{1}{2}\sqrt{2}$ since the times differ exactly one period with $t=0,75$.

d

$t=0,75+k\cdot 6\vee t=2,25+k\cdot 6$.

Exercise 3
a

$h=45+1,5\cdot cos\left(60°\right)=45,75$ m.

b

Smooth graph through $h\left(0\right)=46,5$, $h\left(60\right)=45,75$, $h\left(90\right)=45$, $h\left(120\right)=44,25$, $h\left(180\right)=43,5$, $h\left(240\right)=44,25$, $h\left(270\right)=45$, $h\left(300\right)=45,75$ and $h\left(360\right)=46,5$, etc...

c

$h=45+1,5cos\left(\alpha \right)=46$ gives $\alpha \approx 48,2°$. The asked for distance is $2\cdot sin\left(48,2°\right)\approx 1,49$ m.

Exercise 4
a

$h\left(0\right)=5$ and $h\left(0,5\right)=3,75$.

b

The period is $2$ seconds.

c

$h\left(6\right)=h\left(0\right)=5$ and $h\left(6,5\right)=h\left(0,5\right)=3,75$.

d

$h\left(15\right)=h\left(-1\right)=0$ and $h\left(15,5\right)=h\left(-0,5\right)=3,75$.

e

formulate your own answer.