Find the cubic function of the form y=ax^3 + bx^2 + cx + d which has a relative maximum point at (0,2) and a point of inflection at (-1,-2). How do I even start this one?
First of all, both given points are on the curve, so
for
(0,2) ---> 2 = 0+0+0+d
so we know d=2
(-1,-2) ---> -2 = a(-1) + b(1) + c(1) + d
-a + b + c + 2 = -2
a - b - c = 4
y' = 3ax^2 + 2bx + c
since (0,2) is a max, we know y' = 0 when x = 0
0 = 3a(0) + 2b(0) + c
c = 0
y'' = 6ax + 2b
we know y'' = 0 when x = -1
0 = -6a + 2b
2b = 6a
b = 3a
so in a-b-c=4
a - 3a - 0 = 4
-2a = 4
a = -2 and b=-6 , c=0 , d=2
then the function is
y = -2x^3 - 6x^2 + 2
check my arithmetic.
To find the cubic function of the given form, we can use the information about the relative maximum point and the point of inflection.
Step 1: Relative maximum point at (0,2)
Since the relative maximum point is (0,2), we know that the graph of the cubic function has a maximum value at x = 0. This means that the derivative of the cubic function is zero at x = 0 since it changes from increasing to decreasing.
Step 2: Point of inflection at (-1,-2)
The point of inflection is at (-1,-2), which means that the graph changes concavity at x = -1. This occurs when the second derivative of the cubic function is zero and changes sign.
Step 3: Determine the derivatives
Let's start by finding the first and second derivatives of the cubic function.
The first derivative is obtained by differentiating the cubic function:
y' = 3ax^2 + 2bx + c
The second derivative is obtained by differentiating the first derivative:
y'' = 6ax + 2b
Step 4: Use the given points to form equations
Using the information we have, we can set up two equations based on the derivatives.
Equation 1: For the relative maximum point (0,2)
Since the relative maximum point is (0,2), the derivative y' should be zero when x = 0.
Substituting the values into the equation:
0 = 3a(0)^2 + 2b(0) + c
0 = c
Equation 2: For the point of inflection (-1,-2)
Since the point of inflection is (-1,-2), the second derivative y'' should be zero when x = -1.
Substituting the values into the equation:
0 = 6a(-1) + 2b
-6a = 2b
Step 5: Solve the system of equations
We have two equations:
Equation 1: 0 = c
Equation 2: -6a = 2b
From Equation 1, we know that c = 0. Substituting this into Equation 2:
-6a = 2b
-3a = b
Step 6: Write the cubic function
Now we can write the cubic function in the form y = ax^3 + bx^2 + cx + d using the information from the equations.
Since c = 0, the equation becomes:
y = ax^3 + bx^2 + d
Using the value of b in terms of a (from Step 5):
y = ax^3 + (-3a)x^2 + d
Step 7: Substitute the point (0,2)
Using the point (0,2), we can substitute the values and solve for d.
2 = a(0)^3 + (-3a)(0)^2 + d
2 = d
So the cubic function is:
y = ax^3 - 3ax^2 + 2
Hence, the cubic function of the given form with a relative maximum point at (0,2) and a point of inflection at (-1,-2) is y = ax^3 - 3ax^2 + 2.
To find the cubic function of the given form, we can start by using the information provided about the relative maximum point and the point of inflection.
Let's begin by considering the relative maximum point at (0,2). This point tells us that the cubic function passes through the point (0,2) and has a maximum value of y=2 at x=0.
Since the function has a relative maximum point at (0,2), we know that the coefficient of the x^3 term (a) is negative. Thus, we can start by assuming that a is a negative number.
Next, let's consider the point of inflection at (-1,-2). This point tells us that the function passes through (-1,-2) and changes concavity at that point.
To find the equation of the cubic function, we need to determine the values of coefficients a, b, c, and d.
1. Start with the equation y = ax^3 + bx^2 + cx + d.
2. Since a < 0 (from the relative maximum point), we can let a = -k, where k is a positive number.
3. Now, substitute the values of x and y for the relative maximum point (0,2). We get:
2 = 0 + 0 + 0 + d
d = 2
4. Now, differentiate the equation y = -kx^3 + bx^2 + cx + 2 to find the equation for the derivative. The derivative represents the gradient or slope of the function:
y' = -3kx^2 + 2bx + c
5. Substitute the values of x and y for the point of inflection (-1,-2) and equate the equation for the derivative to zero because the slope is zero at the point of inflection:
0 = -3k(-1)^2 + 2b(-1) + c
0 = 3k - 2b + c (equation 1)
6. We have two equations now: d = 2 and 0 = 3k - 2b + c. We need one more equation to solve for the coefficients. We can use the fact that the derivative at the relative maximum point is zero. Since the function has a maximum at x=0, the derivative at x=0 must equal zero:
0 = -3k(0)^2 + 2b(0) + c
0 = c (equation 2)
7. Now, substitute the value of c = 0 from equation 2 into equation 1:
0 = 3k - 2b
2b = 3k
b = (3k)/2
8. Finally, substitute the values of b and c (from equations 6 and 7) into equation 1 to find the value of k:
0 = 3k - 2((3k)/2)
0 = 3k - 3k
3k = 3k
k can be any positive number
9. Since k can be any positive number, we can choose k = 1 for simplicity.
10. Substitute the values of a = -k, b = (3k)/2, c = 0, and d = 2 into the equation y = ax^3 + bx^2 + cx + d to get the cubic function:
y = -x^3 + (3/2)x^2 + 2
Therefore, the cubic function satisfying the given conditions is y = -x^3 + (3/2)x^2 + 2.