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To find the relative maxima and minima of the given function \( f(x) = 4x^3 - 3x^2 - 36x + 17 \), we follow these steps:

### 1. **Find the first derivative**:
The first derivative of a function helps identify critical points where maxima and minima occur.

\[
f'(x) = \frac{d}{dx}(4x^3 - 3x^2 - 36x + 17)
\]

Applying the power rule:

\[
f'(x) = 12x^2 - 6x - 36
\]

### 2. **Set the first derivative equal to zero to find critical points**:
\[
12x^2 - 6x - 36 = 0
\]

Divide through by 6:

\[
2x^2 - x - 6 = 0
\]

Now solve this quadratic equation using the quadratic formula:

\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-6)}}{2(2)}
\]

\[
x = \frac{1 \pm \sqrt{1 + 48}}{4} = \frac{1 \pm \sqrt{49}}{4} = \frac{1 \pm 7}{4}
\]

Thus, we have two solutions:

\[
x = \frac{1 + 7}{4} = 2 \quad \text{and} \quad x = \frac{1 - 7}{4} = -\frac{3}{2}
\]

So, the critical points are \( x = 2 \) and \( x = -\frac{3}{2} \).

### 3. **Find the second derivative to determine concavity**:
The second derivative will help us determine whether these critical points correspond to maxima or minima.

\[
f''(x) = \frac{d}{dx}(12x^2 - 6x - 36) = 24x - 6
\]

Now evaluate the second derivative at the critical points:

- For \( x = 2 \):

\[
f''(2) = 24(2) - 6 = 48 - 6 = 42 \quad (\text{positive, so } x = 2 \text{ is a minimum})
\]

- For \( x = -\frac{3}{2} \):

\[
f''\left(-\frac{3}{2}\right) = 24\left(-\frac{3}{2}\right) - 6 = -36 - 6 = -42 \quad (\text{negative, so } x = -\frac{3}{2} \text{ is a maximum})
\]

### 4. **Find the corresponding function values**:

- For \( x = 2 \):

\[
f(2) = 4(2)^3 - 3(2)^2 - 36(2) + 17 = 32 - 12 - 72 + 17 = -35
\]

- For \( x = -\frac{3}{2} \):

\[
f\left(-\frac{3}{2}\right) = 4\left(-\frac{3}{2}\right)^3 - 3\left(-\frac{3}{2}\right)^2 - 36\left(-\frac{3}{2}\right) + 17
\]
\[
= 4\left(-\frac{27}{8}\right) - 3\left(\frac{9}{4}\right) + 54 + 17 = -\frac{27}{2} - \frac{27}{4} + 54 + 17 = \frac{-108 - 27 + 432 + 136}{4} = \frac{433}{4} = 108.25
\]

### **Conclusion**:
- There is a relative maximum at \( \left(-\frac{3}{2}, 108.25\right) \).
- There is a relative minimum at \( (2, -35) \).

Would you like a more detailed explanation or have any questions?

Here are five related questions:
1. How can we confirm whether a point is a maximum or minimum using the first derivative test?
2. Can a function have more than one relative maximum or minimum?
3. How is the second derivative test different from the first derivative test?
4. What are the implications of the concavity at a critical point?
5. What would happen if the second derivative equals zero at a critical point?

**Tip**: If the second derivative test is inconclusive, use the first derivative test to analyze the sign changes around the critical points.

Find all relative maxima and minima of the following function: f(x) = 4x^3 - 3x^2 - 36x + 17. Identify the points of relative maximum and minimum, and state the corresponding values of the function at these points.

Learn how to find the relative maxima and minima of the cubic function f(x) = 4x^3 - 3x^2 - 36x + 17 using derivatives. This problem covers the identification of critical points and the application of the second derivative test. Suitable for students in Grades 11-12.

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