Prove that the function given by is increasing in R.
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Prove that the function given by f (x) = x³ – 3x²+ 3x – 100 is increasing in R.
solution :- we know, function f(x) in interval [a,b] is increasing only when f'(x) > 0 in [a,b] .
f(x) = x³ - 3x² + 3x - 100
differentiate f(x) with respect to x
f'(x) = 3x² - 6x + 3 = 3(x² - 2x + 1)
f'(x) = 3(x - 1)²
we know, all (x - 1)² is positive for all real value of x .
so, f'(x) = 3(x - 1)² > for all real value of x
hence, f(x) is increasing for all real value of x
or, f(x) is increasing on R
solution :- we know, function f(x) in interval [a,b] is increasing only when f'(x) > 0 in [a,b] .
f(x) = x³ - 3x² + 3x - 100
differentiate f(x) with respect to x
f'(x) = 3x² - 6x + 3 = 3(x² - 2x + 1)
f'(x) = 3(x - 1)²
we know, all (x - 1)² is positive for all real value of x .
so, f'(x) = 3(x - 1)² > for all real value of x
hence, f(x) is increasing for all real value of x
or, f(x) is increasing on R
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