Asked by david
A function is defined by f(x) = ax^3 + bx^2 + cx + d. Find a, b, c, and d if f(x) has a point of inflection at (0, 2) and a local maximum at (2,6).
Answers
Answered by
Reiny
y = ax^3 + bx^2 + cx + d
(2,6) lies on it, so 6 = 8a + 4b + 2c + d
(0,2) lies on it, so 2 = 0+0+0+d ----> d = 2
then 8a + 4b + 2c = 4
4a + 2b + c = 2 **
y' = 3ax^2 + 2bx + c
since (2,6) is a max
when x = 2, y' = 0
3a(4) + 2b(2) + c = 0
12a + 4b + c = 0 ***
subtract ** from ***
8a + 2b = -2
4a + b = -1
y'' = 6ax + 2b
(0,2) is a point of inflection, so when x = 0, y'' = 0
6a(0) + 2b = 0
2b = 0
b = 0
then in 4a + b = -1, a = -1/4
in ** , 4a + 2b + c = 2
4(-1/4) + 0 + c = 2
c = 3
f(x) = (-1/4)x^3 + 3x + 2
check: https://www.wolframalpha.com/input/?i=plot+f%28x%29+%3D+%28-1%2F4%29x%5E3+%2B+3x+%2B+2
Yup!
(2,6) lies on it, so 6 = 8a + 4b + 2c + d
(0,2) lies on it, so 2 = 0+0+0+d ----> d = 2
then 8a + 4b + 2c = 4
4a + 2b + c = 2 **
y' = 3ax^2 + 2bx + c
since (2,6) is a max
when x = 2, y' = 0
3a(4) + 2b(2) + c = 0
12a + 4b + c = 0 ***
subtract ** from ***
8a + 2b = -2
4a + b = -1
y'' = 6ax + 2b
(0,2) is a point of inflection, so when x = 0, y'' = 0
6a(0) + 2b = 0
2b = 0
b = 0
then in 4a + b = -1, a = -1/4
in ** , 4a + 2b + c = 2
4(-1/4) + 0 + c = 2
c = 3
f(x) = (-1/4)x^3 + 3x + 2
check: https://www.wolframalpha.com/input/?i=plot+f%28x%29+%3D+%28-1%2F4%29x%5E3+%2B+3x+%2B+2
Yup!
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