# Thread: Points of Inflexion on a Trigonometric Graph

1. ## Points of Inflexion on a Trigonometric Graph

I'm having some trouble with this question, I was hoping someone could help me out. The question asks;

Find all points of inflexion on the curve $\displaystyle y = 3\cos(2x + \frac{\pi}{4})$.

2. ## Ok

Originally Posted by Flay
I'm having some trouble with this question, I was hoping someone could help me out. The question asks;

Find all points of inflexion on the curve $\displaystyle y = 3\cos(2x + \frac{\pi}{4})$.

$\displaystyle f'(x)=-6sin\bigg(2x+\frac{\pi}{4}\bigg)$...$\displaystyle f''(x)=-12cos\bigg(2x+\frac{\pi}{4}\bigg)$...no intervals so I am going to assume it is $\displaystyle [0,2\pi]$ the solutions are $\displaystyle x=.3926,x=1.9635,x=3.534,x=5.105$...now using those set up test intervals and test an element of each in f''(x) where it changes signs there is an inflection point
3. Sorry, forgot to mention the interval. You were right in assuming $\displaystyle [0, 2\pi]$. What I don't get about this question, though, is exactly how you solve $\displaystyle f''(x)=-12cos\bigg(2x+\frac{\pi}{4}\bigg)$ to get those values. Could you explain, please?
Sorry, forgot to mention the interval. You were right in assuming $\displaystyle [0, 2\pi]$. What I don't get about this question, though, is exactly how you solve $\displaystyle f''(x)=-12cos\bigg(2x+\frac{\pi}{4}\bigg)$ to get those values. Could you explain, please?
...there are two ways you need to solve this $\displaystyle -12cos\bigg(2x+\frac{\pi}{4}\bigg)=0$...first way $\displaystyle cos\bigg(2x+\frac{\pi}{4}\bigg)=0,2x+\frac{\pi}{4} =\bigg(arccos(0)=\frac{\pi}{2}\bigg),x=\frac{\pi}{ 8}$ then using the trig identites find the other values(e.g. add $\displaystyle \frac{\pi}{2}$)...secondly graph it on your calculator and find the points of intersection between [tex][0,2\pi]]/math]