Question

G'(x)=0,если g(x)=4x+8/x

Answer 1

Ответ:

g'(x) =

$4 - \frac{8}{ {x}^{2} } = 0 = > \\ \frac{8}{ {x}^{2} } = 4 \\ 4 {x}^{2} = 8 \\ {x}^{2} = 2 \\ x = + - \sqrt{2}$

Answer 2

Ответ:

g'(x) =  

4 -  \frac{8}{ {x}^{2} }  = 0  =  >  \\  \frac{8}{ {x}^{2} }  = 4 \\ 4 {x}^{2}  = 8 \\  {x}^{2} =  2 \\ x =  +  -  \sqrt{2}  

Объяснение:

Как-то так.

Answer 3

To find the critical points of the function $g(x) = \frac{4x+8}{x}$, we need to determine the values of $x$ for which the derivative $g'(x)$ is equal to zero.

First, let's find the derivative $g'(x)$ of the function $g(x)$:

$g(x) = \frac{4x + 8}{x}$

Using the quotient rule to differentiate, we get:

$g'(x) = \frac{(x \cdot 0) - (4x + 8) \cdot 1}{x^2} = \frac{-4x - 8}{x^2}$

Next, we set $g'(x) = 0$ to find the critical points:

$\frac{-4x - 8}{x^2} = 0$

To solve for $x$, we can multiply both sides by $x^2$ to get rid of the denominator:

$-4x - 8 = 0$

Now, isolate $x$ on one side:

$4x = -8$

$x = \frac{-8}{4}$

$x = -2$

So, the critical point of the function $g(x) = \frac{4x+8}{x}$ occurs at $x = -2$.