We are asked to find the maximum and minimum values of the function $f(x) = x^2 - 4x + 3$ on the interval $[0, 3]$.

Step 1: Find the critical points.

To find the critical points, we need to compute the derivative of $f(x)$ and set it equal to zero.

$f'(x) = \frac{d}{dx} (x^2 - 4x + 3) = 2x - 4$

Now, set $f'(x) = 0$:

$2x - 4 = 0 \quad \Rightarrow \quad x = 2$

Thus, $x = 2$ is a critical point.

Step 2: Evaluate $f(x)$ at the critical point and at the endpoints of the interval.

We evaluate the function at $x = 0$, $x = 3$, and the critical point $x = 2$:

$f(0) = 0^2 - 4(0) + 3 = 3$ $f(3) = 3^2 - 4(3) + 3 = 9 - 12 + 3 = 0$ $f(2) = 2^2 - 4(2) + 3 = 4 - 8 + 3 = -1$

Step 3: Identify the maximum and minimum values.

• $f(0) = 3$
• $f(2) = -1$
• $f(3) = 0$

Thus, the minimum value of $f(x)$ on the interval $[0, 3]$ is $-1$, which occurs at $x = 2$, and the maximum value is $3$, which occurs at $x = 0$.

Conclusion:

• Minimum value: $-1$ at $x = 2$
• Maximum value: $3$ at $x = 0$

Would you like further details on any of these steps?

Here are 5 related questions for practice:

1. How do you find the derivative of a polynomial function?
2. What does a critical point represent in a function?
3. How do you determine if a critical point is a maximum or minimum?
4. How do you evaluate the function at boundary points of an interval?
5. Can a function have more than one critical point within a given interval?

Tip: Always remember to check both the critical points and the boundary points when finding extrema on a closed interval.