# Question 5baa9

Feb 10, 2017

Given
$y = f \left(x\right) = 2 \sin \left(x - \frac{\pi}{4}\right)$

To get x-intercepts we can put y=0 and solve for $x \in \left[0 , 4 \pi\right]$

So $2 \sin \left(x - \frac{\pi}{4}\right) = 0$

$\implies x - \frac{\pi}{4} = n \pi \text{ where } n \in \mathbb{Z}$

$\implies x = n \pi + \frac{\pi}{4} \text{ where } n \in \mathbb{Z}$

Putting $n = 0$ we get

$x = \frac{\pi}{4}$

Putting $n = 1$ we get

$x = \frac{5 \pi}{4}$

Putting $n = 2$ we get

$x = \frac{9 \pi}{4}$

Putting $n = 3$ we get

$x = \frac{13 \pi}{4}$

(a) So points of x-intercepts over $\left[0 , 4 \pi\right]$ are

(pi/4,0);((5pi)/4,0);((9pi)/4,0);((13pi)/4,0)#

b) For maximum value of y the value of $\sin \left(x - \frac{\pi}{4}\right) = 1$

So $x - \frac{\pi}{4} = \frac{\pi}{2}$
$\implies x = \frac{\pi}{4} + \frac{\pi}{2} = \frac{3 \pi}{4}$

For $x = \frac{3 \pi}{4} \text{ } y = 2$

The point(in (0,2pi) where the graph of this function reaches
maximum is $\left(\frac{3 \pi}{4} , 2\right)$