Question

f(x)=x³+3ax²+3bx+c has a maximum at x=-1 and minimum zero at x=1. Find a, b and c.

Answer 1

f(x) = x³ + 3ax² + 3bx + c has maximum at x = -1 and minimum zero at x = 1

first differentiate f(x) with respect to x,

f'(x) = 3x² + 3ax + 3b

again, differentiate with respect to x,

f"(x) = 6x + 3a .

f(x) will be maximum at x = -1, only when f''(x) < 0 and f'(x) = 0 at x = -1

so, f"(-1) = 6(-1) + 3a < 0 ⇒ a < 2 ....(1)

f'(-1) = 3(-1)² + 3a(-1) + 3b = 0

⇒3 - 3a + 3b = 0

⇒a - b = 1 .....(2)

f(x) will be minimum at x = 1 only when f"(x) > 0 and f'(x) = 0

so, f''(1) = 6 + 3a > 0 ⇒a > -2.....(3)

f'(1) = 3 + 3a + 3b = 0

a + b = -1 ......(4)

from equations (2) and (4),

2a = 0 ⇒a = 0 and b = -1

here, also a satisfied inequalities (1) and (3),

it is given minimum value of f(x) is zero.

i.e., f(1) = (1)³ + 3a(1)² + 3b(1) + c = 0

3a + 3b + c = 0

putting a = 0 and b = -1

so, c = 3

hence, a = 0, b = -1 and c = 3