Prove that f(x)=x³-3x²+3x+100 is increasing on R.
Answers
Theorem: Let f be a differentiable real function defined on an open interval (a,b).
- (i) If f’(x) > 0 for all x ∈ (a, b), then f(x) is increasing on (a, b).
- (ii) If f’(x) < 0 for all x ∈ (a, b), then f(x) is decreasing on (a, b).
here function, f(x) = x³ - 3x² + 3x + 100
differentiating f(x) with respect to x,
⇒f'(x) = 3x² - 6x + 3
⇒f'(x) = 3(x² - 2x + 1) = 3(x - 1)²
we know, square of any function always will be positive for all real value of x.
so, f'(x) = 3(x - 1)² > 0 for all x ∈ R
from above theorem, it is clear that f(x) is increasing on R.
Answer:
Step-by-step explanation:
Theorem: Let f be a differentiable real function defined on an open interval (a,b).
(i) If f’(x) > 0 for all x ∈ (a, b), then f(x) is increasing on (a, b).
(ii) If f’(x) < 0 for all x ∈ (a, b), then f(x) is decreasing on (a, b).
here function, f(x) = x³ - 3x² + 3x + 100
differentiating f(x) with respect to x,
⇒f'(x) = 3x² - 6x + 3
⇒f'(x) = 3(x² - 2x + 1) = 3(x - 1)²
we know, square of any function always will be positive for all real value of x.
so, f'(x) = 3(x - 1)² > 0 for all x ∈ R
from above theorem, it is clear that f(x) is increasing on R.