Question

Найти наименьшее значение функции. y=√x^2-8x+80

Answer 1

Наименьшее значение y=8

Answer 2

To find the minimum value of the function $y = \sqrt{x^2 - 8x + 80}$, we need to determine the critical points. The critical points occur where the derivative of the function is equal to zero or is undefined. After finding the critical points, we can check which one corresponds to the minimum value of the function.

Let's start by finding the derivative of the function:

$y = \sqrt{x^2 - 8x + 80}$

To make the calculations easier, let's rewrite the function as follows:

$y = (x^2 - 8x + 80)^{\frac{1}{2}}$

Now, we'll use the chain rule to find the derivative:

$\frac{dy}{dx} = \frac{1}{2} \cdot (x^2 - 8x + 80)^{-\frac{1}{2}} \cdot (2x - 8)$

Next, we'll set the derivative equal to zero and solve for $x$:

$\frac{dy}{dx} = 0$

$\frac{1}{2} \cdot (x^2 - 8x + 80)^{-\frac{1}{2}} \cdot (2x - 8) = 0$

Now, let's solve for $x$:

$2x - 8 = 0$

$2x = 8$

$x = 4$

Now that we have the critical point $x = 4$, we can find the corresponding $y$ value by plugging it back into the original function:

$y = \sqrt{(4)^2 - 8(4) + 80}$

$y = \sqrt{16 - 32 + 80}$

$y = \sqrt{64}$

$y = 8$

Therefore, the minimum value of the function $y = \sqrt{x^2 - 8x + 80}$ occurs when $x = 4$, and the minimum value is $y = 8$.