ample 6. Find the extreme value of the function $$ y=x^{3}-6 x^{2}+9 x-8 $$

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To find the extreme value of the function $$ y=x^{3}-6x^{2}+9x-8 $$, we need to find the critical points of the function. Critical points are the points where the derivative of the function is equal to zero or undefined.

Let's find the derivative of the function with respect to $$ x $$ and then set it equal to zero to find the critical points.

The derivative of the function $$ y=x^{3}-6x^{2}+9x-8 $$ with respect to $$ x $$ is:

$$ \frac{dy}{dx} = 3x^{2} - 12x + 9 $$

Now, we set the derivative equal to zero to find the critical points:

$$ 3x^{2} - 12x + 9 = 0 $$

Let's solve this quadratic equation to find the critical points.
Solve the quadratic equation by following steps:
- step0: Solve by factoring:
  $$3x^{2}-12x+9=0$$
- step1: Factor the expression:
  $$3\left(x-3\right)\left(x-1\right)=0$$
- step2: Divide the terms:
  $$\left(x-3\right)\left(x-1\right)=0$$
- step3: Separate into possible cases:
  $$\begin{align}&x-3=0\&x-1=0\end{align}$$
- step4: Solve the equation:
  $$\begin{align}&x=3\&x=1\end{align}$$
- step5: Rewrite:
  $$x_{1}=1,x_{2}=3$$
The critical points of the function are $$ x=1 $$ and $$ x=3 $$.

Now, we need to determine whether these critical points correspond to a maximum or minimum value of the function. We can do this by evaluating the second derivative of the function at these critical points.

The second derivative of the function $$ y=x^{3}-6x^{2}+9x-8 $$ with respect to $$ x $$ is:

$$ \frac{d^{2}y}{dx^{2}} = 6x - 12 $$

Evaluating the second derivative at the critical points:

At $$ x=1 $$:
$$ 6(1) - 12 = -6 $$

At $$ x=3 $$:
$$ 6(3) - 12 = 6 $$

Since the second derivative is negative at $$ x=1 $$, this critical point corresponds to a maximum value of the function. Since the second derivative is positive at $$ x=3 $$, this critical point corresponds to a minimum value of the function.

Therefore, the extreme value of the function $$ y=x^{3}-6x^{2}+9x-8 $$ is a maximum value at $$ x=1 $$ and a minimum value at $$ x=3 $$.
The function $$ y=x^{3}-6x^{2}+9x-8 $$ has a maximum at $$ x=1 $$ and a minimum at $$ x=3 $$.

ample 6. Find the extreme value of the function \( y=x^{3}-6 x^{2}+9 x-8 \)

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