Question

Найти наименьшее значение функции y = 6 - 1/(x^2+1)

Answer 1

y=-4/x- x

(-4/x- x)'=0

4/х^2-1=0

4/х^2=1

x^2=4

x=2 и x=-2

2 непринадлежит [-2,5;-1]

f(-2,5)=-4/(-2,5)+2,5=4,1

f(-2)=-4/(-2)+2=4

f(-1)=-4/(-1)+1=5

Ответ: максимальная точка = 5; минимальная точка = 4.

Answer 2

To find the minimum value of the function $y = 6 - \frac{1}{x^2 + 1}$, we need to determine the critical points of the function and then check which one of them corresponds to the minimum.

Step 1: Find the derivative of the function $y$ with respect to $x$. $y = 6 - \frac{1}{x^2 + 1}$

Let's differentiate $y$ with respect to $x$: $\frac{dy}{dx} = 0 + \frac{d}{dx} \left(-\frac{1}{x^2 + 1}\right)$

To differentiate $-\frac{1}{x^2 + 1}$, we can use the chain rule: $\frac{d}{dx} \left(-\frac{1}{x^2 + 1}\right) = -\frac{d}{dx}(x^2 + 1)^{-1}$

The chain rule states that if $u$ is a function of $x$ and $v$ is a function of $u$, then the derivative of $v$ with respect to $x$ is $v' = v \cdot u'$. Applying this to our case where $v = -1$ and $u = (x^2 + 1)^{-1}$:

$\frac{d}{dx} \left(-\frac{1}{x^2 + 1}\right) = -1 \cdot \frac{d}{dx}(x^2 + 1)^{-1}$ $\frac{d}{dx} \left(-\frac{1}{x^2 + 1}\right) = -\frac{1}{(x^2 + 1)^2} \cdot \frac{d}{dx}(x^2 + 1)$ $\frac{d}{dx} \left(-\frac{1}{x^2 + 1}\right) = -\frac{1}{(x^2 + 1)^2} \cdot 2x$

So, the derivative of $y$ with respect to $x$ is: $\frac{dy}{dx} = -\frac{2x}{(x^2 + 1)^2}$

Step 2: Find the critical points. Critical points occur when the derivative is equal to zero or undefined. Since the derivative $\frac{dy}{dx}$ is always defined, the critical points occur when the numerator $2x$ is equal to zero. $2x = 0 \Rightarrow x = 0$

Step 3: Check for the minimum. To determine whether the critical point corresponds to a minimum or maximum, we can use the second derivative test. If the second derivative is positive at the critical point, it is a local minimum. If the second derivative is negative, it is a local maximum. If the second derivative is zero, the test is inconclusive.

Let's find the second derivative of $y$ with respect to $x$: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{2x}{(x^2 + 1)^2}\right)$

To differentiate $-\frac{2x}{(x^2 + 1)^2}$