Question

show that the function x³+3x²+3x+7 is an increasing function for all real value of x.

Answer 1

concept :- any function f(x) is an increasing function in the interval [a, b] only when , f'(x) ≥ 0 in [a , b]


y = x³ + 3x² + 3x + 7
differentiate with respect to x
dy/dx = 3x² + 6x + 3
=3{ x² + 2x + 1}
= 3(x + 1)²

dy/dx = 3(x + 1)² ,it's always positive for all x belongs to real numbers .
hence, function y = x³ + 3x² + 3x + 7 is an increasing function for all x€ R