Application of Derivatives
f (x) = 2 x3 – 6 x2 + 6 x + 5
∴ f ' (x) = 6 x2 – 12 x + 6 = 6 (x2 – 2 x + 1)
∴ f ' (x) = 6 ( x – 1)2
Now f ' (x) = 0 ⇒ 6 (x – 1)2 = 0 ⇒ r = 1
When x < 1 slightly, f ' (x) = 6 (1) = + ve
When x > 1 slightly, f ' (x) = 6 (+) = ve
∴ f ' (x) does not change sign as x passes through I
∴ x = 1 is a point of inflexion.
Sponsor Area
The radius of a circle is increasing uniformly at the rate of 4 cm per second Find the rate at which the area of the circle is increasing when the radius is 8 cm.
Sponsor Area
Sponsor Area