Question

Найти наибольшее и наименьшее значение функции y=cos^4x-sin^4x

Answer 1

(сos² x - sin² x)(сos² x + sin² x)= сos² x - sin² x=cos 2x

наибольшее значение +1 ,

а наименьшее -1.

Answer 2

To find the maximum and minimum values of the function $y = \cos^4(x) - \sin^4(x)$, we need to take the derivative of the function with respect to $x$ and then solve for the critical points.

Let's start by finding the derivative of $y$ with respect to $x$:

$\frac{dy}{dx} = 4\cos^3(x)(-\sin(x)) - 4\sin^3(x)\cos(x)$

Now, to find the critical points, we need to solve for $x$ when $\frac{dy}{dx} = 0$:

$4\cos^3(x)(-\sin(x)) - 4\sin^3(x)\cos(x) = 0$

Factor out $4$:

$4\cos(x)\sin(x)(\cos^2(x) - \sin^2(x)) = 0$

Now, we have three cases to consider:

1. $\cos(x) = 0$ (Case 1)
2. $\sin(x) = 0$ (Case 2)
3. $\cos^2(x) - \sin^2(x) = 0$ (Case 3)

Solving for each case:

Case 1: $\cos(x) = 0$

This occurs at $x = \frac{\pi}{2} + k\pi$, where $k$ is an integer.

Case 2: $\sin(x) = 0$

This occurs at $x = k\pi$, where $k$ is an integer.

Case 3: $\cos^2(x) - \sin^2(x) = 0$

Using the trigonometric identity $\cos^2(x) - \sin^2(x) = \cos(2x)$:

$\cos(2x) = 0$

This occurs at $x = \frac{\pi}{4} + \frac{k\pi}{2}$, where $k$ is an integer.

Now, we have all the critical points. We need to evaluate $y$ at each of these points and also at the endpoints of the interval to determine the maximum and minimum values.

Let's consider the interval $[0, 2\pi]$:

1. $x = 0$: $y = \cos^4(0) - \sin^4(0) = 1 - 0 = 1$
2. $x = \frac{\pi}{2}$: $y = \cos^4\left(\frac{\pi}{2}\right) - \sin^4\left(\frac{\pi}{2}\right) = 0 - 1 = -1$
3. $x = \pi$: $y = \cos^4(\pi) - \sin^4(\pi) = 1 - 0 = 1$
4. $x = \frac{3\pi}{2}$: $y = \cos^4\left(\frac{3\pi}{2}\right) - \sin^4\left(\frac{3\pi}{2}\right) = 0 - 1 = -1$
5. $x = 2\pi$: $y = \cos^4(2\pi) - \sin^4(2\pi) = 1 - 0 = 1$

Now, let's consider the critical points:

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