Question

Find the values of a for which f(x)=ax³-9ax²+9x+25 is increasing on R

Answer 1

function, f(x) = ax³ - 9ax² + 9x + 25

differentiating f(x) with respect to x,

f'(x) = d(ax³ - 9ax² + 9x + 25)/dx

= a d(x³)/dx - 9a d(x²)/dx + 9 dx/dx + d(25)/dx

= a × 3x² - 9a × 2x + 9 + 0

= 3ax² - 18ax + 9

= 3(ax² - 6ax + 3)

a/c to question,

f(x) is increasing on R.

so, f'(x) > 0 on R

i.e., f'(x) = 3(ax² - 6ax + 3) > 0 on R.

⇒ax² - 6ax + 3 > 0 This is possible only if coefficient of x² > 0 and discriminant < 0

i.e., a > 0 and D = (-6a)² - 4(a)(3) < 0

D = 36a² - 12a < 0

⇒12a(3a - 1) < 0

⇒0 < a < 1/3

hence, a ∈ (0, 1/3)