Question

Y=x-cosx x=-pi/2.pi/2 экстремумы

Answer 1

Находим производную:

$y'=(x-\cos x)'=x'-(\cos x)'=1+\sin x$

Поскольку при всех x выполнено неравенство $-1\leqslant \sin x\leqslant 1$, то всегда $y'\geqslant 1+(-1)=0$. Если производная принимает только неотрицательные значения, то функция (возможно, нестрого) возрастает, минимальные значения на отрезке принимает в левом конце отрезка, максимальные – в правом.

$\displaystyle\min\limits_{x\in\left[-\frac\pi2,\frac\pi2\right]}y(x)=y\left(-\frac\pi2\right)=-\frac\pi2-\cos\left(-\frac\pi2\right)=-\frac\pi2-0=-\frac\pi2$

$\displaystyle\max\limits_{x\in\left[-\frac\pi2,\frac\pi2\right]}y(x)=y\left(\frac\pi2\right)=\frac\pi2-\cos\left(-\frac\pi2\right)=\frac\pi2-0=\frac\pi2$

Answer 2

To find the extrema of the function $y = x - \cos x$ over the interval $x = -\frac{\pi}{2}$ to $x = \frac{\pi}{2}$, we need to locate the critical points and determine if they are local maxima or minima. The critical points occur where the derivative of the function is equal to zero or is undefined.

First, let's find the derivative of the function $y = x - \cos x$:

$y' = \frac{dy}{dx} = 1 + \sin x$

Now, let's set the derivative equal to zero and solve for $x$ to find the critical points:

$1 + \sin x = 0$

$\sin x = -1$

The solutions to $\sin x = -1$ in the interval $[- \frac{\pi}{2}, \frac{\pi}{2}]$ are $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$.

Now, we need to check the second derivative to determine the nature of these critical points. The second derivative is:

$y'' = \frac{d^2y}{dx^2} = \cos x$

Evaluating the second derivative at $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$:

$y''(-\frac{\pi}{2}) = \cos(-\frac{\pi}{2}) = 0$

$y''(\frac{\pi}{2}) = \cos(\frac{\pi}{2}) = 0$

Since the second derivative is zero at both critical points, the test is inconclusive, and we need to use another method to determine the nature of the critical points.

One way to do this is to examine the behavior of the function near the critical points. We can make a small table of values to see how the function behaves around $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$:

For $x$ values close to $-\frac{\pi}{2}$:

$$\begin{array}{|c|c|} \hline x & x - \cos x \\ \hline -\frac{\pi}{2} - 0.1 & -0.095\\ -\frac{\pi}{2} - 0.01 & -0.00995\\ -\frac{\pi}{2} + 0.01 & 0.01005\\ -\frac{\pi}{2} + 0.1 & 0.105\\ \hline \end{array}$$

For $x$ values close to $\frac{\pi}{2}$:

$$\begin{array}{|c|c|} \hline x & x - \cos x \\ \hline \frac{\pi}{2} - 0.1 & 0.105\\ \frac{\pi}{2} - 0.01 & 0.00995\\ \frac{\pi}{2} + 0.01 & -0.01005\\ \frac{\pi}{2} + 0.1 & -0.095\\ \hline \end{array}$$

From the table, we can see that for values of $x$ slightly less than $-\frac{\pi}{2}$, the function is increasing, and for values of $x$ slightly greater than 0 0