Question

Let f(x) be a cubic polynomial with f(1) = –10, f(–1) = 6, and has a local minima at x = 1, and f'(x) has a local minima at x = –1. Then f(3) is equal to _____.

Answer 1

let f(x) = ax³ + bx² + cx + d

\(\because\) and (1) = -10 and f(-1) = 6

\(\therefore\) a + b + c + d = -10.....(1)

and -a + b - c + d = 6....(2)

\(\because\) f(x) has a local minima at x = 1

\(\therefore\)f'(1) = 0

and f'(x) has a local minima at x = -1

\(\therefore\) f" (-1) = 0

\(\because\) f(x) = ax³ + bx² + cx + d

\(\therefore\) f'(x) = 3ax² + 2bx + c

f" (x) = 6ax + 2b

\(\because\) f" (-1) = 0

= -6a + 2b = 0

= b = 3a...(3)

also f'(1) = 0

= 3a + 2b + c = 0

= c = -9a....(4)

By adding (1) and (2) , we get

2b + 2d = -4

= b + d = -2

= 3a + d = -2

= d = - 2 - 3a....(5)

Put b = 3a,c = -9a and d = -2-3a in (1) we get

a + 3a - 9a - 2 - 3a = -10

= -8a = -10 + 2 = -8

a = \(\cfrac{-8}{-8}\) = 1

\(\therefore\) b = 3, c = -9 and d = - 2 - 3 = - 5

\(\therefore\) f(x) = x³ + 3x² - 9x - 5

\(\therefore\)f (3) = 3³ + 3.3² - 9 x 3 - 5

= 27 + 27 - 27 - 5

= 27 - 5

= 22

Answer 2

F'(x) = a(x – 1)(x + 3)

F"(x) = 6a(x + 1)

F'(x) = 3a(x + 1)² + b

F'(1) = 0

\(\Rightarrow\) b = –12a

F(x) = a(x + 1)³ – 12ax + c

= (x + 1)³ – 12x – 6

F(3) = 64 – 36 – 6 = 22