I'm desperate.. I keep getting b=0 and I know it's wrong!!
Find a cubic function f(x) = ax^3 + bx^2 + cx + d that has a local maximum at (-3,3) and a local minimum at (2,0) knowing that c=-12b and d=3-27b.
You can reduce the number of unkowns to 2 (a and b) by rewriting the equatiosn with substitutions for c and d.
The system appears overdetermined. You could write four equations in two unknowns knowing the value of two points and the fact that f'(x) = 0 at those points. I would try to use the facts that
f'(x) = 3ax^2 + 2bx -12b = 0 at x = -3 and x=2 to solve for a and b.
27a -6b -12b = 27a -18b = 0
12a +4b -12b = 12a -8b = 0
Those last two equations are equivalent; you can only use one of them. So you need to use an equation that says f(-2) = 0, for example.
8a + 4b -[12b*(-2)] +3 -27b = 0
8a + b = -3
24a + 3b = -9
24a -16b = 0
19 b = -9
This does not lead to an integer value for b. I may have made a mistake somewhere. Check my work and thinking
I understand everything up until the last part with the using equations. Where did 8a and 4b come from?
I too got myself all caught up in nasty fractions, so I tried a different approach.
Since (2,0) is a minimum and it touches the x-axis, there has to be a double root at x=2
so we have (x-2)(x-2) as factors
since it is a cubic, there can be only one other linear factor
so let the function be
y = a(x-b)(x-2)^2 , my b is not necessarily the same as the drwls's b)
I then found the derivative of that and subbed in the fact that if x=-3, dy/dx = 0
this led to 55a + 10ab = 0
a(55 + 10b) = 0
a = 0, not possible if we want a cubic
or
b = -11/2
I then subbed in (-3,3) in y = a(x-b)(x-2)^2 and using the fact that b=-11/2 I got
a = 6/125
so my function is
f(x) = (6/125)(x+11/2)(x-2)^2
both of your points (2,0) and (-3,3) satisfy,
I then found the derivative of that function and set it equal to zero
that gave me x = 2, or x = -3
Q.E.D.
I will leave it up to your to expand it.
As drwls also noted, there seems to be redundant information.
To find the cubic function that satisfies the given conditions, we can start by using the information about the local maximum and minimum points.
A local maximum occurs when the derivative changes from positive to negative, and a local minimum occurs when the derivative changes from negative to positive. So, we will need to find the derivative of the cubic function.
Let's proceed step by step:
Step 1: Find the derivative of the function f(x).
The derivative of a cubic function f(x) = ax^3 + bx^2 + cx + d can be obtained by differentiating each term with respect to x. Since we have the values of c and d in terms of b, we can rewrite the function as follows: f(x) = ax^3 + bx^2 - 12bx + (3-27b).
The derivative of f(x) is:
f'(x) = 3ax^2 + 2bx - 12b.
Step 2: Use the information about the local maximum and minimum.
At the local maximum point (-3, 3), the derivative is equal to zero. Substituting x = -3 into f'(x) = 3ax^2 + 2bx - 12b, we get:
3a(-3)^2 + 2b(-3) - 12b = 0.
Simplifying this equation gives: 9a - 6b - 12b = 0.
Similarly, at the local minimum point (2, 0), the derivative is also equal to zero. Substituting x = 2 into f'(x) = 3ax^2 + 2bx - 12b, we get:
3a(2)^2 + 2b(2) - 12b = 0.
Simplifying this equation gives: 12a + 10b - 12b = 0.
Step 3: Solve the system of equations.
We now have a system of two equations with two unknowns:
9a - 6b - 12b = 0, (equation 1)
12a + 10b - 12b = 0. (equation 2)
Simplifying equation 1 gives: 9a - 18b = 0.
Dividing by 9, we have: a - 2b = 0. (equation 3)
Simplifying equation 2 gives: 12a - 2b = 0. (equation 4)
Now we have a system of two linear equations:
a - 2b = 0, (equation 3)
12a - 2b = 0. (equation 4)
Step 4: Solve the system of equations to find the values of a and b.
By comparing equations 3 and 4, we can see that they are essentially the same equation.
Combining equations 3 and 4 gives: a - 2b = 12a - 2b.
Simplifying this equation, we have: -11a = 0.
Dividing by -11, we get: a = 0.
Substituting the value of a into equation 3, we have: 0 - 2b = 0.
Simplifying this equation gives: -2b = 0.
Dividing by -2, we get: b = 0.
Step 5: Substitute the values of a and b into c and d.
Since we have c = -12b and d = 3 - 27b, substituting b = 0 into these equations gives us:
c = -12(0) = 0,
d = 3 - 27(0) = 3.
Step 6: Write the cubic function.
Now that we have the values for a, b, c, and d, we can write the cubic function f(x) = ax^3 + bx^2 + cx + d as:
f(x) = 0x^3 + 0x^2 + 0x + 3.
Simplifying this, we get:
f(x) = 3.
So, the cubic function that has a local maximum at (-3,3) and a local minimum at (2,0) is f(x) = 3.