Find a, b, c, and d such that the cubic function ax^3 + bx^2 + cx + d satisfies the given conditions
Relative maximum: (2,4)
Relative minimum: (4,2)
Inflection point: (3,3)
So this is what I have so far:
f'(x) = 3a^2 + 2bx + c
f''(x) = 6ax + 2b
f(2) = a(8) + b(4) + c(2) + d= 4
f(4) = a(64) + b(16) + c(4) + d = 2
--> 30a + 6b + c = -1
Am I going in the right direction? If I am, what do I do next? Thanks.
your thinking is ok so far
also remember that (3,3) also lies on the original function, so
27a + 9b + 3c + d = 3 will be another equation.
How did you get 30a in 30a + 6b + c = -1
If you subtracted your two equations and then divided by 2 should it not have been
28a + 6b + c = -1 ? ---> #4
by subtracting my new equation from 64a +16b+4c+d = 2 , I get a new equation
37a + 7b+c=-1 ---> # 5
#5 - #4 :
9a + b = 0
b = -9a
also we know that f'(2) = 0 and f'(4) = 0
f'(2) = 12a + 4b + c = 0
f'(4) = 48a + 8b + c = 0
subtract:
36a + 4b = 0
b = -9a , nothing new here , confirms above.
but we also know f'(2) = f''(3) = 0
12a + 4b + c = 18a + 2b
c = 6a - 2b
c = 6a - 2(-9a) = 24a
in #4:
28a + 6(-9a) + 24a = -1
-2a = -1
a = 1/2
b= -9/2
c = 12
so now back into : 8a + 4b + 2c + d = 4
4 -18 +24 + d = 4
d = -6
so f(x) = (1/2)x^3 -(9/2)x^2 + 12x - 6
check my arithmetic
You have a good start, but there are a few corrections to make:
f'(x) = 3a(x)^2 + 2bx + c
f''(x) = 6ax + 2b
f(2) = a(8) + b(4) + c(2) + d= 4
f(4) = a(64) + b(16) + c(4) + d = 2
--> 30a + 6b + c = -1
You have three conditions for max. at (2,4):
f(2)=4
f'(2)=0
f"(2)<0
relative min. at f(4)=2 =>
f(4)=2
f'(4)=0
f"(4)>0
Inflexion point at (3,3) means:
f(3)=3
f"(3)=0
So you'd have six equations for four unknowns a,b, c and d. Notice that if the system is consistent (which is the case here), you only need 4 of the conditions to find the solutions, but you will have to check that the remaining conditions are satisfied.
Can you take it from here?
Well, it looks like you're on the right track, but let me lend you a hand (or a funny bone) in the right direction!
Given the conditions, let's start with the derivative and second derivative of the cubic function:
f'(x) = 3ax^2 + 2bx + c
f''(x) = 6ax + 2b
To find the values of a, b, c, and d, we can use the relative maximum, relative minimum, and inflection point.
1. Relative maximum: (2, 4)
At the relative maximum, the derivative is 0. So, plug in x = 2 into f'(x) and set it equal to 0:
3a(2)^2 + 2b(2) + c = 0
12a + 4b + c = 0
2. Relative minimum: (4, 2)
At the relative minimum, the derivative is 0 again. Plug in x = 4 into f'(x) and set it equal to 0:
3a(4)^2 + 2b(4) + c = 0
48a + 8b + c = 0
3. Inflection point: (3, 3)
At the inflection point, the second derivative is 0. So, plug in x = 3 into f''(x) and set it equal to 0:
6a(3) + 2b = 0
18a + 2b = 0
Now you have a system of three equations with three variables (a, b, c). Solve the system and find the values of a, b, and c. Once you have those, you can then find d by plugging in one of the given points, for example, (2, 4), into the cubic function.
Remember, solving equations can be a bit complex, but it's nothing we can't solve with a pinch of humor and some mathematical wizardry!
Yes, you are going in the right direction. To find the values of a, b, c, and d, we need to solve the system of equations obtained from the given conditions and the derivatives of the cubic function.
1. Relative maximum: (2,4)
At x=2, the derivative f'(x) should be equal to zero.
This gives us the equation: 3a(2)^2 + 2b(2) + c = 0 --> 12a + 4b + c = 0
2. Relative minimum: (4,2)
At x=4, the derivative f'(x) should be equal to zero.
This gives us the equation: 3a(4)^2 + 2b(4) + c = 0 --> 48a + 8b + c = 0
3. Inflection point: (3,3)
The second derivative f''(x) should be equal to zero at x=3.
This gives us the equation: 6a(3) + 2b = 0 --> 18a + 2b = 0
4. f(2) = 4
This gives us the equation: a(2)^3 + b(2)^2 + c(2) + d = 4 --> 8a + 4b + 2c + d = 4
5. f(4) = 2
This gives us the equation: a(4)^3 + b(4)^2 + c(4) + d = 2 --> 64a + 16b + 4c + d = 2
Now we have a system of 5 equations with 4 variables (a, b, c, d). We can solve this system to find the values of a, b, c, and d.
Yes, you're on the right track! To find the values of a, b, c, and d, we need to set up a system of equations using the given information.
First, we know that the relative maximum occurs at (2, 4). To find this, we need to find the derivative of the cubic function and set it equal to zero:
f'(x) = 3ax^2 + 2bx + c
To find the derivative at x = 2, substitute 2 into the equation:
0 = 3a(2)^2 + 2b(2) + c
0 = 12a + 4b + c (Equation 1)
Next, we know that the relative minimum occurs at (4, 2). Similar to the previous step, substitute 4 into the equation:
0 = 3a(4)^2 + 2b(4) + c
0 = 48a + 8b + c (Equation 2)
Finally, we know that the inflection point occurs at (3, 3). To find this, we need to find the second derivative of the cubic function and set it equal to zero:
f''(x) = 6ax + 2b
To find the second derivative at x = 3, substitute 3 into the equation:
0 = 6a(3) + 2b
0 = 18a + 2b (Equation 3)
Now, we have three equations (Equations 1, 2, and 3) and three unknowns (a, b, c). We can solve this system of equations to find the values of a, b, and c.
You've already gotten the equation from the given condition f(2) = 4:
a(8) + b(4) + c(2) + d = 4 (Equation 4)
To summarize, we have the following system of equations:
Equation 1: 12a + 4b + c = 0
Equation 2: 48a + 8b + c = 0
Equation 3: 18a + 2b = 0
Equation 4: 8a + 4b + 2c + d = 4
Now you can solve this system of linear equations to find the values of a, b, c, and d.