Show that `f(x)=x^(3)-6x^(2)+18x+5` is an increasing function for all `x in R`, Findits value when the rate increases of f(x) is least.

Answer

To show that the function \( f(x) = x^3 - 6x^2 + 18x + 5 \) is an increasing function for all \( x \in \mathbb{R} \) and to find its value when the rate of increase is least, we will follow these steps:

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

The first derivative of the function \( f(x) \) is calculated as follows:

\[ f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 18x + 5) \]

Calculating the derivatives term by term:

- The derivative of \( x^3 \) is \( 3x^2 \).
- The derivative of \( -6x^2 \) is \( -12x \).
- The derivative of \( 18x \) is \( 18 \).
- The derivative of the constant \( 5 \) is \( 0 \).

Combining these results, we get:

\[ f'(x) = 3x^2 - 12x + 18 \]

### Step 2: Analyze the first derivative

To show that \( f(x) \) is increasing for all \( x \in \mathbb{R} \), we need to prove that \( f'(x) > 0 \) for all \( x \).

We can rewrite the first derivative:

\[ f'(x) = 3(x^2 - 4x + 6) \]

### Step 3: Determine the nature of the quadratic \( x^2 - 4x + 6 \)

Next, we analyze the quadratic \( x^2 - 4x + 6 \). To determine if it is always positive, we calculate its discriminant:

\[ D = b^2 - 4ac = (-4)^2 - 4 \cdot 1 \cdot 6 = 16 - 24 = -8 \]

Since the discriminant \( D < 0 \), the quadratic \( x^2 - 4x + 6 \) has no real roots and opens upwards (as the coefficient of \( x^2 \) is positive). Therefore, it is positive for all \( x \in \mathbb{R} \):

\[ x^2 - 4x + 6 > 0 \quad \text{for all } x \in \mathbb{R} \]

Thus, we conclude that:

\[ f'(x) = 3(x^2 - 4x + 6) > 0 \quad \text{for all } x \in \mathbb{R} \]

This shows that \( f(x) \) is an increasing function for all \( x \in \mathbb{R} \).

### Step 4: Find the minimum value of \( f'(x) \)

To find when the rate of increase of \( f(x) \) is least, we need to find the critical points of \( f'(x) \). We do this by finding the second derivative \( f''(x) \):

\[ f''(x) = \frac{d}{dx}(f'(x)) = \frac{d}{dx}(3x^2 - 12x + 18) = 6x - 12 \]

Setting the second derivative equal to zero to find critical points:

\[ 6x - 12 = 0 \implies x = 2 \]

### Step 5: Verify that this is a minimum

Now we check the third derivative \( f'''(x) \):

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

Since \( f'''(x) = 6 > 0 \), this confirms that \( f'(x) \) has a minimum at \( x = 2 \).

### Step 6: Calculate \( f(2) \)

Now we find the value of \( f(x) \) at \( x = 2 \):

\[ f(2) = 2^3 - 6(2^2) + 18(2) + 5 \]

Calculating each term:

- \( 2^3 = 8 \)
- \( -6(2^2) = -6(4) = -24 \)
- \( 18(2) = 36 \)
- \( +5 = 5 \)

Combining these:

\[ f(2) = 8 - 24 + 36 + 5 = 25 \]

### Conclusion

Thus, the function \( f(x) \) is an increasing function for all \( x \in \mathbb{R} \), and the value of \( f(x) \) when the rate of increase is least is:

\[ \boxed{25} \]