Find a cubic function y=ax^3+bx^2+cx+d that has a local maximum value of 3 at x=-2 and a local minimum value of 0 at x=1
To find the cubic function that satisfies the given conditions, we can use the local maximum and minimum values to determine the coefficients of the function.
Given:
Local maximum: y = 3 at x = -2
Local minimum: y = 0 at x = 1
1. Start by considering the general form of a cubic function: y = ax^3 + bx^2 + cx + d.
2. Use the local maximum value to find the value of "a":
Since the local maximum occurs at x = -2, substitute these values into the function:
3 = a(-2)^3 + b(-2)^2 + c(-2) + d
Simplify the equation:
3 = -8a + 4b - 2c + d (equation 1)
3. Use the local minimum value to find the value of "a":
Since the local minimum occurs at x = 1, substitute these values into the function:
0 = a(1)^3 + b(1)^2 + c(1) + d
Simplify the equation:
0 = a + b + c + d (equation 2)
4. Solve the system of equations (equations 1 and 2) to find the values of "b" and "c":
To solve the system, subtract equation 2 from equation 1:
3 = -8a + 4b - 2c + d
0 = a + b + c + d
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3 = -9a + 3b - 3c (equation 3)
5. Substitute any arbitrary value for "a" and solve for the remaining coefficients:
Choose a value for "a". Let's say a = 1 for simplicity.
Substitute it into equation 3:
3 = -9(1) + 3b - 3c
3 = -9 + 3b - 3c
3b - 3c = 12 (equation 4)
6. Choose another arbitrary value for "b" or "c" and solve for the remaining coefficient:
Let's assume b = 1, which makes the equation 4:
3(1) - 3c = 12
3 - 3c = 12
-3c = 9
c = -3
7. Substitute the values of "a", "b", and "c" into equation 2 to find "d":
0 = (1) + (1) + (-3) + d
0 = -1 + d
d = 1
8. Plug the values of "a", "b", "c", and "d" back into the general form of the cubic function:
y = ax^3 + bx^2 + cx + d
y = 1x^3 + 1x^2 + (-3)x + 1
y = x^3 + x^2 - 3x + 1
Therefore, the cubic function that satisfies the given conditions is:
y = x^3 + x^2 - 3x + 1.