Find the maximum and minimum of f(*x*) = *x* 2 − 4 *x* +3 on the interval [0, 3].

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.

Find the maximum and minimum of f(x) = x^2 − 4x + 3 on the interval [0, 3].

Explore how to find the maximum and minimum values of the quadratic function f(x) = x^2 − 4x + 3 on the interval [0, 3]. Learn how to compute derivatives, identify critical points, and apply the Extreme Value Theorem. This detailed solution is suitable for students in Grades 11-12.

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