Find a cubic function f(x) = 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.
f ' (x) = 3ax^2 + 2bx + c
f '(-2) = 0
12a -4b + c = 0 (#!1
f ' (1) = 0
3a + 2b + c = 0 (#2)
#1 - #2 ----> 9a - 6b = 0
or 3a = 2b
also (-2,3) lies on original
-8a + 4b - 2c + d = 3
and (1,0) lies on it
a + b + c + d = 0
subtract those two equations
9a -3b+3c = -3 ,or
3a - b + c = -1 , (#3)
#2 - #3 :
3b = 1
b = 1/3 , but a = 2b/3 = (2/3)(1/3) = 2/9
in #3
3(2/9) -1/3 + c = -1
c = -4/3
in a+b+c+d = 0
2/9 + 1/3 - 4/3 + d = 0
d = 7/9
so f(x) = (2/9)x^3 + (1/3)x^2 - (4/3)x + 7/9
That's hard I am working on a simlar question after 9 years
It's very hard
Why did the cubic function go to therapy? Because it was having an identity crisis! Well, let's solve this puzzle, shall we?
To find a cubic function that satisfies these conditions, we can start by finding the x-coordinate of the inflection point (where the function changes concavity). This occurs when the derivative of the function equals zero. So, let's find the derivative of the cubic function:
f(x) = ax^3 + bx^2 + cx + d
f'(x) = 3ax^2 + 2bx + c
Since the derivative equals zero at the inflection point, let's plug in x = -2 and solve:
0 = 3a(-2)^2 + 2b(-2) + c
0 = 12a - 4b + c ...(Equation 1)
Similarly, for the inflection point at x = 1, we have:
0 = 3a(1)^2 + 2b(1) + c
0 = 3a + 2b + c ...(Equation 2)
Now, to find the cubic function that satisfies these conditions, we can set certain values for a, b, and c. Let's choose a = 1, b = 1, and c = -2. By substituting these values into Equations 1 and 2, we can find d:
Equation 1: 0 = 12(1) - 4(1) + c
c = -8
Equation 2: 0 = 3(1) + 2(1) + c
0 = 3 + 2 - 8
0 = -3
Therefore, we have a cubic function with a = 1, b = 1, c = -2, and d = -3:
f(x) = x^3 + x^2 - 2x - 3
Hope this cubic function brings the x-citement you were looking for!
To find a cubic function that satisfies the given conditions, we need to solve a system of equations.
The given information implies the following:
1. f(-2) = 3: This means that when x = -2, the function has a value of 3.
2. f(1) = 0: This means that when x = 1, the function has a value of 0.
Using these conditions, we can set up the following system of equations:
f(-2) = (-2)^3a + (-2)^2b + (-2)c + d = 3
f(1) = (1)^3a + (1)^2b + (1)c + d = 0
To determine a unique solution, we need one additional equation.
In this case, we can use the fact that the function has a local maximum at x = -2 and a local minimum at x = 1.
At a local maximum or minimum point, the derivative of the function is zero. Hence, we can differentiate the cubic function and set it equal to zero to get a third equation.
f'(x) = 3ax^2 + 2bx + c
First, let's differentiate f(x), the cubic function:
f'(x) = 3ax^2 + 2bx + c
Now, set f'(-2) = 0 and f'(1) = 0:
3a(-2)^2 + 2b(-2) + c = 0 -> 12a - 4b + c = 0 -> Equation 1
3a(1)^2 + 2b(1) + c = 0 -> 3a + 2b + c = 0 -> Equation 2
Now we have three equations we can solve simultaneously:
1. (-2)^3a + (-2)^2b + (-2)c + d = 3
2. (1)^3a + (1)^2b + (1)c + d = 0
3. 12a - 4b + c = 0
4. 3a + 2b + c = 0
Using these equations, we can solve for a, b, c, and d to find the cubic function that satisfies the given conditions.
Find the integral of and equation made by multiplying the factors using the roots, using A and B as variables, then, plug in the points to the integral and subtract the two equations to get rid of B, then solve for A. Plug in A to one of the point equations, then, solve for B, then, plug A and B into integral, expand, and BAM!
finding integral: A(x-p)(x-q) + B -> expand and then find integral.
At the local min/max points, the derivative is zero.
f' = 3ax^2 + 2bx + c = 0
Plug in x = -3 and x = 2, and that gives you two equations.
Also, use f = ax^3 + bx^2 + cx + d
Plug in the values (x = -3, f = 3) and (x = 2, f = 0).
That gives you two more equations.
Now solve the four simultaneous linear equations in a, b, c, and d.