Question

Find the values of k for which f(x) = kx3 - 9kx2 + 9x + 3 is increasing on R

Answer 1

f(x ) = kx^3 -9kx^2 +9x +3 is increasing only when f'(x) >0

now differentiate f(x) wrt x
f(x) =3kx^2-18kx+9 >0

kx^2 -6kx +3 >0

this is possible only when,
k >0

and Discriminant = (6k)^2 -4(3)(k) >0
3k-1 > 0
k < 1/3
now, k € (0 , 1/3)
hence, if k €(0 , 1/3) then function f(x) is increasing .

here € means belongs to

Answer 2

f(x) = k x³ - 9 k x² + 9 x + 3   ,   x ∈ R

f '(x) = 3 k x² - 18 k x + 9  = 3 (k x² - 6 k x + 3)

Find the roots of f '(x):   x = [ 3k +- √(9k²-3k) ] / k 
     If the roots are real, then in between the two roots, the value of f '(x) is negative. and at the roots, the derivative is zero.  So roots must be imaginary.
     Then discriminant < 0
     3 k (3 k - 1) < 0
     So    0 < k < 1/3

For this range of values f(x) is always increasing.