We have,
f(x)=x3-3x2+3x-100
f'(x)=3x2-6x+3
=3(x2-2x+1)
=3(x-1)2
For any x∈R, (x − 1)2 > 0.
Thus, f'(x) is always positive in R.
Hence, the given function (f) is increasing in R
What is the Planning Process?
Increasing Function:
On an interval I, a function f(x) is said to be increasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) ≤ f(y)
Decreasing Function:
On an interval I, a function f(x) is said to be decreasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) ≥ f(y)
Strictly Increasing Function:
On an interval I, a function f(x) is said to be strictly increasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) < f(y)
Strictly Decreasing Function:
On an interval I, a function f(x) is said to be strictly decreasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) > f(y)