Let `f(x)=x^3-3x^2+ 3x + 4`, comment on the monotonic behaviour of `f(x)` at `x=0` `x=1`

Answer

To analyze the monotonic behavior of the function \( f(x) = x^3 - 3x^2 + 3x + 4 \) at the points \( x = 0 \) and \( x = 1 \), we will follow these steps:

### Step 1: Find the derivative of \( f(x) \)

To determine the monotonic behavior of the function, we first need to find its derivative \( f'(x) \).

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

Calculating the derivative term by term:

- The derivative of \( x^3 \) is \( 3x^2 \).
- The derivative of \( -3x^2 \) is \( -6x \).
- The derivative of \( 3x \) is \( 3 \).
- The derivative of the constant \( 4 \) is \( 0 \).

Combining these results, we get:

\[ f'(x) = 3x^2 - 6x + 3 \]

### Step 2: Simplify the derivative

We can factor the derivative:

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

### Step 3: Evaluate the derivative at \( x = 0 \)

Now, we will evaluate \( f'(x) \) at \( x = 0 \):

\[ f'(0) = 3(0 - 1)^2 = 3(1) = 3 \]

### Step 4: Evaluate the derivative at \( x = 1 \)

Next, we will evaluate \( f'(x) \) at \( x = 1 \):

\[ f'(1) = 3(1 - 1)^2 = 3(0) = 0 \]

### Step 5: Analyze the monotonic behavior

- At \( x = 0 \):
- Since \( f'(0) = 3 > 0 \), the function \( f(x) \) is **increasing** at \( x = 0 \).

- At \( x = 1 \):
- Since \( f'(1) = 0 \), the function \( f(x) \) is **neither increasing nor decreasing** at \( x = 1\).

### Conclusion

- At \( x = 0 \), \( f(x) \) is increasing.
- At \( x = 1 \), \( f(x) \) is neither increasing nor decreasing.