Let f(x) `=x^(3) -x^(2) -x-4`
(i) find the possible points of maxima `//`minima of f(x) ` x in R`
(ii) Find the number of critical points of f(x) ` x in [1,3]`
(iii) Discuss absolute (global) maxima `//` minima value of f(x) for `x in [-2,2]`
(iv) Prove that for `x in (1,3)` the function does not has a Global maximum.

Answer

To solve the problem step by step, let's break it down according to the parts of the question.

### Given Function:
Let \( f(x) = x^3 - x^2 - x - 4 \).

### (i) Find the possible points of maxima/minima of \( f(x) \) for \( x \in \mathbb{R} \).

**Step 1:** Find the first derivative \( f'(x) \).

\[ f'(x) = \frac{d}{dx}(x^3 - x^2 - x - 4) = 3x^2 - 2x - 1 \]

**Step 2:** Set the first derivative equal to zero to find critical points.

\[ 3x^2 - 2x - 1 = 0 \]

**Step 3:** Factor or use the quadratic formula to solve for \( x \).

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} = \frac{2 \pm \sqrt{4 + 12}}{6} = \frac{2 \pm \sqrt{16}}{6} = \frac{2 \pm 4}{6} \]

This gives us:

\[ x = 1 \quad \text{and} \quad x = -\frac{1}{3} \]

**Step 4:** Determine whether these points are maxima or minima by using the second derivative test.

\[ f''(x) = \frac{d}{dx}(3x^2 - 2x - 1) = 6x - 2 \]

Evaluate \( f''(x) \) at the critical points:

- For \( x = 1 \):
\[ f''(1) = 6(1) - 2 = 4 > 0 \quad \text{(local minimum)} \]

- For \( x = -\frac{1}{3} \):
\[ f''\left(-\frac{1}{3}\right) = 6\left(-\frac{1}{3}\right) - 2 = -2 - 2 = -4 < 0 \quad \text{(local maximum)} \]

**Conclusion for (i):**
- Local maximum at \( x = -\frac{1}{3} \)
- Local minimum at \( x = 1 \)

### (ii) Find the number of critical points of \( f(x) \) for \( x \in [1, 3] \).

**Step 1:** From part (i), we found critical points at \( x = -\frac{1}{3} \) and \( x = 1 \).

**Step 2:** Check if these points lie within the interval \( [1, 3] \).

- The critical point \( x = -\frac{1}{3} \) is not in the interval.
- The critical point \( x = 1 \) is in the interval.

**Conclusion for (ii):** There is **1 critical point** in the interval \( [1, 3] \).

### (iii) Discuss absolute (global) maxima/minima value of \( f(x) \) for \( x \in [-2, 2] \).

**Step 1:** Evaluate \( f(x) \) at the endpoints of the interval and at the critical point found in part (i).

- \( f(-2) = (-2)^3 - (-2)^2 - (-2) - 4 = -8 - 4 + 2 - 4 = -14 \)
- \( f(2) = (2)^3 - (2)^2 - (2) - 4 = 8 - 4 - 2 - 4 = -2 \)
- \( f(1) = (1)^3 - (1)^2 - (1) - 4 = 1 - 1 - 1 - 4 = -5 \)

**Step 2:** Compare the values:

- \( f(-2) = -14 \)
- \( f(1) = -5 \)
- \( f(2) = -2 \)

**Conclusion for (iii):**
- Absolute minimum is \( -14 \) at \( x = -2 \).
- Absolute maximum is \( -2 \) at \( x = 2 \).

### (iv) Prove that for \( x \in (1, 3) \), the function does not have a global maximum.

**Step 1:** From part (i), the local maximum occurs at \( x = -\frac{1}{3} \), which is not in the interval \( (1, 3) \).

**Step 2:** The only critical point in \( (1, 3) \) is \( x = 1 \), which is a local minimum.

**Step 3:** Since the function is continuous and differentiable in \( (1, 3) \), and the only critical point is a minimum, there cannot be a global maximum.

**Conclusion for (iv):** The function does not have a global maximum in the interval \( (1, 3) \).