Correct Answer - Option 1 : f(x) is an increasing function on R

**Concept**:

Let f(x) be a function defined on an interval (a, b), this function is said to be an increasing function:

- If x
_{1} < x_{2} then f(x_{1}) ≤ f(x_{2}) ∀ x_{1}, x_{2} ∈ (a, b)
- Here, \(\frac{{dy}}{{dx}} \ge 0\;or\;f'\left( x \right) \ge 0\)

Similarly, f(x) is said to be a decreasing function:

- If If x1 < x2 then f(x
_{1}) ≥ f(x_{2}) ∀ x_{1}, x_{2} ∈ (a, b).
- Here, \(\frac{{dy}}{{dx}} \le 0\;or\;f'\left( x \right) \le 0\)

**Calculation**:

Given: f(x) = x3 - 6x2 + 12x - 18

Let's find out f'(x)

⇒ f'(x) = 3x^{2} - 12x + 12 = 3 ⋅ (x^{2} - 4x + 4)

⇒ f'(x) = 3 ⋅ (x - 2)^{2}

Now as we know that, for any x ∈ R we have (x - 2)^{2} ≥ 0

⇒ f'(x) ≥ 0

As we know that for an increasing function say f(x) we have f'(x) ≥ 0

Hence, the given function f(x) is an increasing function on R.