Sol.f(x) =

⇒ f ′ (x) =

+ +

⇒ f ′ (x) = (x + b 2 ) (x + c 2 ) –b 2 c 2 + (x + a 2 ) (x + c 2 ) –a 2 c 2 + (x + a 2 ) (x + b 2 ) –a 2 b 2

⇒ f ′ (x) = 3x 2 + 2x (a 2 + b 2 + c 2 )

For f (x) to be increasing, we must have

f ′ (x) > 0

⇒ 3x 2 + 2x (a 2 + b 2 + c 2 ) > 0

⇒ x {3x + 2(a 2 + b 2 + c 2 )} > 0

⇒ x < – (a 2 + b 2 + c 2 ) or x > 0

⇒ x ∈ ∪ (0, ∞ )

So, f(x) is increasing on

∪ (0, ∞ )

For f(x) to be decreasing, we must have f ′ (x) < 0

⇒ 3x 2 + 2x (a 2 + b 2 + c 2 ) < 0

⇒ x(3x + 2(a 2 + b 2 + c 2 ) ) < 0

⇒ – (a 2 + b 2 + c 2 ) < x < 0 [See fig. below]

⇒ x ∈

So, f(x) is decreasing on

Hence, f(x) is increasing on

∪ (0, ∞ ) and

decreasing on