If the graph of cubic function f(x)=ax^3+bx^2+cx+d?

Question

has a local max at (0,6) and a local min at (2,2), compute the value of f(1)

Answer 1

use horner table to simplize and then its easy

Answer 2

f(x) = ax^3 + bx^2 + cx + d
f'(x) = 3ax^2 + 2bx + c

using the four points given, you can find a, b, c and d:
6 = a(0)^3 + b(0)^2 + c(0) + d = d
2 = a(2)^3 + b(2)^2 + c(2) + d = 8a + 4b + 2c + d
0 = 3a(0)^2 + 2b(0) + c = c
0 = 3a(2)^2 + 2b(2) + c = 12a + 4b + c

now, notice that c = 0 and d = 6; use this to get:
2 = 8a + 4b + 6
0 = 12a + 4b
or
8a + 4b = -4
12a + 4b = 0 (or 4b = -12a)

8a - 12a = -4
-4a = -4
a = 1

12a + 4b = 0
12(1) = -4b
12 = -4b
b = -3

now you get a=1, b=-3, c=0, d=6
f(x) = x^3 - 3x^2 + 6
f(1) = (1)^3 - 3(1)^2 + 6 = 1 - 3 + 6 = 4

Answer 3

Function:

f(x)=ax^3+bx^2+cx+d

First derivative function:

f'(x) = 3ax^2 + 2bx + c

Coefficient a, b, c and d can be find from the given points of local maximums.

Solve:

f(0) = 6
f(2) = 2
f'(0) = 0
f'(2) = 0

d = 6
8a + 4b + 2c + d = 2
c = 0
12a + 4b + c = 0

Continue please, I am lazy.