2 votes

Study of a trigonometric function.

I have doubt with this exercise because I have not clear when is increasing the function g (x) = cos ^ 2 (x). [Drifting I have to _ * cos (x) = 0 * _ and I don't know at what interval is increased if x \[0,pi\]

2voto

Anon User Points 0
:) Hello! Bluenile Rain. It seems to me that you were wrong when deriving the function in study. Let's see:! [] (// blob.todoexpertos.com/uploads/md/4201fcdc6c95017d6eef1fe1efa94377.png) (You can click on the image to enlarge it) :)

1voto

Anon User Points 0
If the function is in the interval \ [0, pi \] Can you take as a reference to calculate, the intervals of the function cos ?, that is, if the cos function passes through 0, pi / 2 and pi, Can you calculate in pi / 4 and 3pi / 4?

1voto

Anon User Points 0
We have the function `
$$\begin{align}&g(x)=\cos^2(2x)\end{align}$$`In this case the function **_g_** * * * is increasing in a certain interval I (or intervals) when g'(I) > 0 (strictly increasing) or g'(I) ≥ 0 (increasing). Then to find such intervals:

`
$$\begin{align}&\dfrac{dg}{dx}=\dfrac{d}{dx}\cos^2(2x)\\&\\&\dfrac{dg}{dx}=\dfrac{d}{d(2x)}\cos^2(2x)\cdot \dfrac{d(2x)}{dx}\\&\\&\dfrac{dg}{dx}=2\cdot\dfrac{d}{d(2x)}\cos^2(2x)\\&\\&\dfrac{dg}{dx}=2\cdot\dfrac{d}{d[\cos(2x)]}\cos^2(2x)\cdot \dfrac{d[\cos(2x)]}{d(2x)}\\&\\&\dfrac{dg}{dx}=2\cdot[2\cos(2x)]\cdot[-\sin(2x)]\\&\\&\dfrac{dg}{dx}=-4\sin(2x)\cos(2x)\\&\\&\dfrac{dg}{dx}= -2\sin(4x)\\&\\&\\&\text{Calculating the interval of growth ...}\\&\\&\dfrac{dg}{dx}\geq 0\\&\\&-2\sin(4x)\geq 0\\&\\&\sin(4x)\leq 0\\&\\&\text{As we know, the function $\sin $ is negative when its argument is in the IIIC and IVC}\\&\text{i.e. }4x\in[\pi,2\pi]\cup[3\pi,4\pi]\cup\cdots \cup[(2k-1)\pi,2k\pi]\cup\cdots=\bigcup\limits_{k\in\mathbb{N}}[(2k-1)\pi,2k\pi]\\&\\&\text{on the other hand the function $\sin$ is negative when the argument is negative, and this belongs to IC and IIC}\\&\\&\text{that is to say }4x\in[-\pi,0]\cup[-3\pi,-2\pi]\cup\cdots\cup[-(2k-1)\pi,-2(k-1)\pi]\cup\cdots=\bigcup\limits_{k\in\mathbb{N}}[-(2k-1)\pi,-2(k-1)\pi]\\&\\&\text{Finally }4x\in \bigcup\limits_{k\in\mathbb{N}}[(2k-1)\pi,2k\pi]\cup[-(2k-1)\pi,-2(k-1)\pi]\\&\\&x\in \bigcup\limits_{k\in\mathbb{N}}\left[\frac{2k-1}{4}\pi,\frac{k}{2}\pi\right]\cup\left[-\frac{2k-1}{4}\pi,-\frac{k-1}{2}\pi\right]\end{align}$$`

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