Find the absolute extrema of the function on the closed interval. $$ y=3 x^{2} \gamma^{3}-2 x,[-1,1] $$ minimum $$ \left.(x, y)=\left(\frac{1,3}{}\right) x\right) $$ maximum $$ (x, y)=\left(\frac{1,1)}{}\right) $$

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We are given the function
$$
y = 3x^2 - 2x
$$
on the closed interval $$[-1,1]$$. We wish to find the absolute minimum and maximum values on this interval.

**Step 1. Find the critical points.**

Differentiate the function:
$$
y' = \frac{d}{dx}(3x^2-2x)=6x-2.
$$
Set the derivative equal to zero:
$$
6x-2=0 \quad\Longrightarrow\quad x=\frac{2}{6}=\frac{1}{3}.
$$
Since $$\frac{1}{3}$$ is in the interval $$[-1,1]$$, it is a candidate for an extremum.

**Step 2. Evaluate the function at the endpoints and at the critical point.**

1. At $$x=-1$$:
   $$
   y(-1)=3(-1)^2-2(-1)=3(1)+2=3+2=5.
   $$
2. At the critical point $$x=\frac{1}{3}$$:
   $$
   y\left(\frac{1}{3}\right)=3\left(\frac{1}{3}\right)^2-2\left(\frac{1}{3}\right)
   =3\left(\frac{1}{9}\right)-\frac{2}{3}
   =\frac{1}{3}-\frac{2}{3}
   =-\frac{1}{3}.
   $$
3. At $$x=1$$:
   $$
   y(1)=3(1)^2-2(1)=3-2=1.
   $$

**Step 3. Determine the absolute extrema.**

- The smallest value is $$-\frac{1}{3}$$ at $$x=\frac{1}{3}$$.
- The largest value is $$5$$ at $$x=-1$$.

Thus, the absolute minimum is 
$$
\left(\frac{1}{3}, -\frac{1}{3}\right)
$$
and the absolute maximum is 
$$
(-1, 5).
$$
The absolute minimum of the function $$ y = 3x^2 - 2x $$ on the interval $$[-1,1]$$ is $$-\frac{1}{3}$$ at $$ x = \frac{1}{3} $$, and the absolute maximum is $$5$$ at $$ x = -1 $$.

Find the absolute extrema of the function on the closed interval.

minimum
maximum

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Find the absolute extrema of the function on the closed interval. minimum maximum