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)
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
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
To determine the constants a, b, c, and d, we need to use the given conditions that the curve has a local maximum at (-2, 4) and a point of inflection at (0, 0).
1. Local Maximum at (-2, 4):
A local maximum occurs at a stationary point, where the derivative of the function is zero. In this case, the derivative of f(x) should be zero at x = -2.
Take the derivative of f(x):
f'(x) = 3ax^2 + 2bx + c
Set f'(-2) = 0:
3a(-2)^2 + 2b(-2) + c = 0
12a - 4b + c = 0 ----(1)
2. Point of Inflection at (0, 0):
To find a point of inflection, we need to find the second derivative of f(x) and set it equal to zero at x = 0.
Take the second derivative of f(x):
f''(x) = 6ax + 2b
Set f''(0) = 0:
6a(0) + 2b = 0
2b = 0
b = 0 ----(2)
Now that we have found the value of b, we can substitute it back into equation (1) to solve for a and c.
12a - 4(0) + c = 0
12a + c = 0 ----(3)
Since f(-2) = 4, substitute x = -2 into f(x) and solve for a, b, and c.
f(-2) = a(-2)^3 + b(-2)^2 + c(-2) + d = 4
-8a + 4c + d = 4 ----(4)
Now we have three equations: (2), (3), and (4), which can be used to solve for a, c, and d.
Solve equations (2) and (3) simultaneously:
Substitute b = 0 into equation (3):
12a + c = 0
c = -12a
Now substitute c = -12a into equation (4):
-8a + 4(-12a) + d = 4
-8a - 48a + d = 4
-56a + d = 4 ----(5)
We now have two equations, (3) and (5), which can be used to solve for a and d.
Solve equations (3) and (5) simultaneously:
Substitute c = -12a from equation (3) into equation (5):
-56a + d = 4
-56a + (-12a) = 4
-68a = 4
a = -4/68
a = -1/17
Now substitute the value of a into equation (3):
12a + c = 0
12(-1/17) + c = 0
-12/17 + c = 0
c = 12/17
Finally, substitute the values of a and c into equation (4):
-8a + 4c + d = 4
(-8)(-1/17) + 4(12/17) + d = 4
8/17 + 48/17 + d = 4
56/17 + d = 4
d = 4 - 56/17
d = (68 - 56)/17
d = 12/17
Therefore, the constants for the cubic function f(x) = ax^3 + bx^2 + cx + d, with a local max at (-2, 4) and a point of inflection at (0, 0), are:
a = -1/17
b = 0
c = 12/17
d = 12/17