# trig transformation

• Apr 28th 2009, 08:18 PM
nikita007
trig transformation
hello, i need help with this particular transformation, I was just wondering if my answer is correct

the transformation is this f(x) = 3 cos 2 (x - π/4) + 1 it asks me to find the period, amplitude, phase shift, which I have as π, 3 and π/4 to the right respectively. It then goes on to ask me to find the first five critical points and this is were I'm slightly confused, i have them as; (π/4, 2), (4π/4, 1), (8π/4, 2), (12π/4, 1), (16π/4, 2). My intuition tells me that these are horribly horribly horribly wrong. Could someone please enlighten me on what I am doing wrong here? I just really want a nudge in the right direction. Thank you very much.
• Apr 29th 2009, 07:49 AM
Hello nikita007

Welcome to Math Help Forum!
Quote:

Originally Posted by nikita007
hello, i need help with this particular transformation, I was just wondering if my answer is correct

the transformation is this f(x) = 3 cos 2 (x - π/4) + 1 it asks me to find the period, amplitude, phase shift, which I have as π, 3 and π/4 to the right respectively.

I agree!

Quote:

It then goes on to ask me to find the first five critical points and this is were I'm slightly confused, i have them as; (π/4, 2), (4π/4, 1), (8π/4, 2), (12π/4, 1), (16π/4, 2). My intuition tells me that these are horribly horribly horribly wrong. Could someone please enlighten me on what I am doing wrong here? I just really want a nudge in the right direction. Thank you very much.
The critical points are where the graph of the function has zero gradient. In the case of sine and cosine functions, these will be where the sine and cosine have their maximum and minimum values of +1 and -1 respectively.

In the function $f(x) = 3\cos2(x-\tfrac{\pi}{4}) + 1$, this will be where:

$\cos2(x-\tfrac{\pi}{4}) = \pm 1$ and $f(x) =\pm 3 + 1 = 4\text{ or }-2$.

Now you know that $\cos\theta = \pm 1$ when $\theta = 0, \pi, 2\pi, ...$ . So work from there, starting with:

$2(x-\tfrac{\pi}{4}) = 0 \Rightarrow x = \tfrac{\pi}{4} \Rightarrow f(x)= 4$

Then $2(x-\tfrac{\pi}{4}) = \pi \Rightarrow x = \tfrac{3\pi}{4}$ and $f(x)= -2$

... and so on.