Find all of the critical points and relative extrema of f(x) = 1 + 8x − 3x^2 

We are tasked with finding the critical points and relative extrema of the function:

\[
f(x) = 1 + 8x - 3x^2
\]

### Step 1: Find the first derivative of \( f(x) \)
To find the critical points, we need to take the first derivative of \( f(x) \) and set it equal to zero:

\[
f'(x) = \frac{d}{dx}(1 + 8x - 3x^2)
\]

\[
f'(x) = 8 - 6x
\]

### Step 2: Set the first derivative equal to zero
Now, set the first derivative equal to zero to find the critical points:

\[
8 - 6x = 0
\]

Solve for \( x \):

\[
6x = 8
\]

\[
x = \frac{8}{6} = \frac{4}{3}
\]

Thus, \( x = \frac{4}{3} \) is the critical point.

### Step 3: Determine the nature of the critical point
To determine whether this critical point is a relative maximum or minimum, we can use the second derivative test.

### Step 4: Find the second derivative of \( f(x) \)
The second derivative of \( f(x) \) is:

\[
f''(x) = \frac{d}{dx}(8 - 6x) = -6
\]

Since \( f''(x) = -6 \), which is negative, the function is concave down at \( x = \frac{4}{3} \). Therefore, the critical point at \( x = \frac{4}{3} \) is a relative maximum.

### Step 5: Calculate the relative maximum value
To find the value of the relative maximum, substitute \( x = \frac{4}{3} \) into the original function:

\[
f\left(\frac{4}{3}\right) = 1 + 8\left(\frac{4}{3}\right) - 3\left(\frac{4}{3}\right)^2
\]

First, simplify the terms:

\[
f\left(\frac{4}{3}\right) = 1 + \frac{32}{3} - 3\left(\frac{16}{9}\right)
\]

\[
f\left(\frac{4}{3}\right) = 1 + \frac{32}{3} - \frac{48}{9}
\]

Find a common denominator:

\[
f\left(\frac{4}{3}\right) = 1 + \frac{32}{3} - \frac{16}{3}
\]

\[
f\left(\frac{4}{3}\right) = 1 + \frac{16}{3}
\]

\[
f\left(\frac{4}{3}\right) = \frac{3}{3} + \frac{16}{3} = \frac{19}{3}
\]

Thus, the relative maximum value is \( \frac{19}{3} \) at \( x = \frac{4}{3} \).

### Conclusion:
- Critical point: \( x = \frac{4}{3} \)
- Relative maximum: \( f\left(\frac{4}{3}\right) = \frac{19}{3} \)

Would you like to go over more details or have any questions? Here are some follow-up questions to explore:

1. What is the second derivative test and how does it work?
2. Can critical points exist without relative extrema?
3. What is the difference between absolute and relative extrema?
4. How would the behavior of the function change if the coefficient of \(x^2\) were positive?
5. What are the implications of a negative second derivative?

**Tip:** The second derivative test is a quick way to determine whether a critical point is a maximum or minimum by checking the concavity of the function at that point.

Find all of the critical points and relative extrema of f(x) = 1 + 8x − 3x^2

Discover how to find the critical points and relative extrema of the function f(x) = 1 + 8x − 3x^2. This detailed solution explains key calculus concepts, including using the first and second derivative tests. Ideal for students in Grades 10-12.

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