Math Snap

STEP 1

Assumptions
1. The function given is $y = \frac{1}{3}x^3 - x + 1$.
2. We need to find the absolute maximum and minimum values of $y$ on the interval $[0, 3]$.
3. The function is continuous on the closed interval $[0, 3]$.

STEP 2

To find the absolute maximum and minimum values of a continuous function on a closed interval, we need to evaluate the function at critical points and at the endpoints of the interval.

STEP 3

First, find the derivative of the function $y = \frac{1}{3}x^3 - x + 1$.
$$y' = \frac{d}{dx} \left(\frac{1}{3}x^3 - x + 1\right)$$

STEP 4

Calculate the derivative.
$$y' = x^2 - 1$$

STEP 5

Find the critical points by setting the derivative equal to zero and solving for $x$.
$$x^2 - 1 = 0$$

STEP 6

Solve the equation $x^2 - 1 = 0$.
$$x^2 = 1$$

STEP 7

Take the square root of both sides to find the critical points.
$$x = \pm 1$$

STEP 8

Since we are only interested in the interval $[0, 3]$, consider only the critical point $x = 1$ (as $x = -1$ is outside the interval).

STEP 9

Evaluate the function $y$ at the critical point $x = 1$.
$$y(1) = \frac{1}{3}(1)^3 - 1 + 1$$

STEP 10

Calculate $y(1)$.
$$y(1) = \frac{1}{3} - 1 + 1 = \frac{1}{3}$$

STEP 11

Evaluate the function $y$ at the endpoints of the interval, $x = 0$ and $x = 3$.
$$y(0) = \frac{1}{3}(0)^3 - 0 + 1$$

STEP 12

Calculate $y(0)$.
$$y(0) = 1$$

STEP 13

Evaluate the function at the other endpoint $x = 3$.
$$y(3) = \frac{1}{3}(3)^3 - 3 + 1$$

STEP 14

Calculate $y(3)$.
$$y(3) = \frac{1}{3}(27) - 3 + 1 = 9 - 3 + 1 = 7$$

STEP 15

Compare the values of $y$ at the critical point and the endpoints: $y(0) = 1$, $y(1) = \frac{1}{3}$, $y(3) = 7$.

Determine the absolute maximum and minimum values.
- The absolute maximum value is $7$ at $x = 3$.
- The absolute minimum value is $\frac{1}{3}$ at $x = 1$.