If `f '(x) phi (x) (x-2) ^(2).` Were `phi(2) ne 0 and phi (x)` is continuous at `x =2` then in the neighbouhood of `x =2`

Options

A. f is increasing if `phi (2) lt 0`
B. f is decreasing if `phi (2) gt0`
C. f is neither increasing nor decreasing
D. f is increasin if `phi(2) gt 0`

Answer

To solve the problem step by step, we need to analyze the given function and its derivative.

### Step 1: Understand the given information
We are given that:
\[ f'(x) = \phi(x) (x - 2)^2 \]
where \(\phi(2) \neq 0\) and \(\phi(x)\) is continuous at \(x = 2\).

### Step 2: Evaluate \(f'(2)\)
Since \((x - 2)^2\) is always non-negative and equals zero at \(x = 2\), we can evaluate \(f'(2)\):
\[ f'(2) = \phi(2) \cdot (2 - 2)^2 = \phi(2) \cdot 0 = 0 \]
Thus, \(f'(2) = 0\).

### Step 3: Analyze the sign of \(f'(x)\) around \(x = 2\)
Next, we need to determine the behavior of \(f'(x)\) in the neighborhood of \(x = 2\).

1. **For \(x < 2\)**:
- \((x - 2)^2 > 0\)
- Therefore, \(f'(x) = \phi(x)(x - 2)^2\) will have the same sign as \(\phi(x)\).

2. **For \(x > 2\)**:
- Again, \((x - 2)^2 > 0\)
- Thus, \(f'(x) = \phi(x)(x - 2)^2\) will also have the same sign as \(\phi(x)\).

### Step 4: Determine the behavior of \(\phi(2)\)
Since \(\phi(2) \neq 0\) and \(\phi(x)\) is continuous at \(x = 2\), we can conclude:
- If \(\phi(2) > 0\), then \(\phi(x) > 0\) for \(x\) near \(2\). Hence, \(f'(x) > 0\) for \(x \neq 2\), indicating that \(f(x)\) is increasing in the neighborhood of \(x = 2\).
- If \(\phi(2) < 0\), then \(\phi(x) < 0\) for \(x\) near \(2\). Hence, \(f'(x) < 0\) for \(x \neq 2\), indicating that \(f(x)\) is decreasing in the neighborhood of \(x = 2\).

### Step 5: Conclusion
From the analysis:
- If \(\phi(2) > 0\), \(f(x)\) is increasing around \(x = 2\).
- If \(\phi(2) < 0\), \(f(x)\) is decreasing around \(x = 2\).

Thus, we conclude that:
- \(f(x)\) is neither strictly increasing nor strictly decreasing at \(x = 2\) since it can change behavior depending on the sign of \(\phi(2)\).