find an expression for a cubic function f if f(1)=6 and f(-1)=f(0)=f(2)=0
from f(0) = 0 and f(2) = 0 and f(-1)=0
we know that there are 3 factors,
x(x-2)(x+1)
Since it is a cubic, there should be a third factor, let it be
ax + b)
so
f(x) = x(x-2)(x+1)
BUT, (1,6) does not satisfy this, so we must include a constant a , so that
f(x) = ax(x-2)(x+1)
for (1,6) to lie on it
6 = a(1)(1-2)(1+1)
6 = -2a
a = -3
f(x) = -3x(x-2)(x+1)
Very good
Why did the cubic function go to therapy? Because it had a lot of zeros and couldn't find its roots!
But don't worry, I'm here to help. Let's find the expression for this cubic function.
Since f(1) = 6, we can start by using the factored form of a cubic function. We know that when x = 1, f(x) = 6. So one of the factors must be (x - 1).
f(x) = (x - 1)(something)
Now, let's find the other two factors. Since f(-1) = f(0) = f(2) = 0, we can set up the equations:
(-1 - 1)(something) = 0
(0 - 1)(something) = 0
(2 - 1)(something) = 0
We can see that the common factor in all three equations is (x - 1).
Therefore, the expression for the cubic function f is:
f(x) = (x - 1)(x + 1)(x - 2)
So, this is the expression for the cubic function given the conditions f(1) = 6 and f(-1) = f(0) = f(2) = 0.
To find an expression for a cubic function f(x) with the given conditions, we can use the fact that the function has zeros at x = -1, x = 0, and x = 2.
A cubic function can be written in factored form as:
f(x) = a(x - r1)(x - r2)(x - r3)
Where a is a constant and r1, r2, and r3 are the roots of the function.
Since f(x) = 0 at x = -1, x = 0, and x = 2, we can substitute these values into the factored form equation:
0 = a(-1 - r1)(-1 - r2)(-1 - r3) ... (1)
0 = a(0 - r1)(0 - r2)(0 - r3) ... (2)
0 = a(2 - r1)(2 - r2)(2 - r3) ... (3)
From equation (2), we can see that the constant a must be zero since one of the factors is multiplied by zero. Therefore, our cubic function can be written as:
f(x) = 0(x - r1)(x - r2)(x - r3)
= 0
So, the expression for the cubic function f(x) is f(x) = 0.
To find an expression for a cubic function, we need to use the given conditions. Let's consider a general cubic function of the form f(x) = ax^3 + bx^2 + cx + d.
According to the given information:
f(1) = 6: Substitute x = 1 into the function and equate it to 6:
a(1)^3 + b(1)^2 + c(1) + d = 6
a + b + c + d = 6 ----(1)
f(-1) = 0: Substitute x = -1 into the function and equate it to 0:
a(-1)^3 + b(-1)^2 + c(-1) + d = 0
-a + b - c + d = 0 ----(2)
f(0) = 0: Substitute x = 0 into the function and equate it to 0:
a(0)^3 + b(0)^2 + c(0) + d = 0
d = 0 ----(3)
f(2) = 0: Substitute x = 2 into the function and equate it to 0:
a(2)^3 + b(2)^2 + c(2) + d = 0
8a + 4b + 2c + d = 0 ----(4)
We will now solve this system of equations to find the values of a, b, c, and d.
From equation (3), we find d = 0.
Substituting d = 0 in equation (1), we get:
a + b + c = 6 ----(5)
Substituting d = 0 in equation (2), we get:
-a + b - c = 0 ----(6)
Substituting d = 0 in equation (4), we get:
8a + 4b + 2c = 0 ----(7)
To solve these equations simultaneously, we can eliminate the variable a by adding equations (5) and (6):
(a + b + c) + (-a + b - c) = 6 + 0
2b = 6
b = 3
Substituting b = 3 into equation (5), we get:
a + 3 + c = 6
a + c = 3 ----(8)
Substituting b = 3 into equation (6), we get:
-a + 3 - c = 0
-a - c = -3 ----(9)
Adding equations (8) and (9), we can eliminate the variable c:
(a + c) + (-a - c) = 3 + (-3)
0 = 0
This tells us that a and c can take any values as long as their sum is 3.
Therefore, the expression for a cubic function f(x) with f(1) = 6 and f(-1) = f(0) = f(2) = 0 is:
f(x) = (a)x^3 + (3 - a)x + 0
Where a and c can be any values as long as their sum is 3.