We know (0,0) lies on it, so d = 0
at(-2,4) --> -8a + 4b - 2c = 4 , #1
In a cubic the point of inflection lies midway between the 2 stationary points, so (2, -4) is the other
at(2,-4) --> 8a +4b + 2c = -4 , #2
add #1 and #2 ------> 8b = 0
b = 0
so #2 becomes 8a + 2c = -4
4a + c = -2 , #3
f ' (x) = 3ax^2 + 2bx + c
at a max/min f'(x) = 0
f'(-2) = 0 = 12a - 4b + c
but we know b = 0, so
12a + c = 0 , #4
subtract #4 - #3
8a = 2
a = 1/4
back in #3,
4(1/4) + c = -2
c = -3
a = 1/4, b = 0 , c = -3, d = 0
the function was:
f(x) = (1/4)x^3 - 3x
verification:
www.wolframalpha.com/input?i=graph+f%28x%29+%3D+%281%2F4%29x%5E3+-+3x+from+-3+to+3
Determine the constants a,b,c and d so that the curve defined by the cubic
f(x) = ax3+bx2+cx+d has a local max at the point (-2,4) and a point of inflection at the point (0,0)
2 answers
f = ax^3 + bx^2 + cx + d
f' = 3ax^2 + 2bx + c
f" = 6ax+2b
so, we need f"(0) = 0, so b=0
f" = 2ax
f' = 3ax^2 + c
f = ax^3 + cx + d
f'(-2) = 0, and f(-2) = 4, and f(0) = 0, so
d = 0
12a+c = 0
-8a - 2c = 4
a = 1/4, c = -3
that leaves us with
f(x) = 1/4 x^3 - 3x
f' = 3ax^2 + 2bx + c
f" = 6ax+2b
so, we need f"(0) = 0, so b=0
f" = 2ax
f' = 3ax^2 + c
f = ax^3 + cx + d
f'(-2) = 0, and f(-2) = 4, and f(0) = 0, so
d = 0
12a+c = 0
-8a - 2c = 4
a = 1/4, c = -3
that leaves us with
f(x) = 1/4 x^3 - 3x