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 $$ (x, y)=\left(\frac{1,1)}{}\right) $$

Question

UpStudy AI · Accepted Answer

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

Solución de inteligencia artificial de Upstudy

Responder

Solución

Extra Insights

preguntas relacionadas

Latest Calculus Questions

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