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प्रश्न
Find intervals of concavity and points of inflexion for the following functions:
f(x) = x(x – 4)3
उत्तर
f'(x) = 3x(x – 4)2 + (x – 4)3(1)
= (x – 4)2 (4x – 4)
= 4 (x – 4)2 (x – 1)
f'(x) = 4 [(x-4)2 (1) + (x – 1)2 (x – 4)]
= 4(x – 4)(x – 4 + 2x – 2)
= 4(x – 4)(3x – 6)
= 12(x – 4)(x – 2)
f'(x) = 0
⇒ 12 (x – 4) (x – 2) = 0
Critical points x = 2, 4
The intervals are `(- oo, 2)`, (2, 4) and `(4, oo)`
In the interval `(-oo, 2)`, f”(x) > 0 ⇒ Curve is Concave upward
In the interval (2, 4), f”(x) < 0 ⇒ Curve is Concave downward.
In the interval `(4, oo)`, f”(x) > 0 ⇒ Curve is Concave upward.
The curve is concave upward in
`(-oo, 2), (4, oo)` it is concave downward in (2, 4).
f”(x) changes its sign when passing through x = 2 and x = 4
Now f(2) = 2(2 – 4)3
= – 16 and f(4)
= 4(4 – 4)3
= 0
∴ The points of inflection are (2, -16) and (4, 0).
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