- #1
Bob123Bob
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- 0
f(x) = ax^3 + bx^2 + cx + d min(1, -4) max(-2, 1) find a,b,c,d
Find a,b,c,d given max and min in cubic function
In summary: So, the equation becomes:a+b+c+d=\frac{10}{27}+\frac{5}{9}+-\frac{20}{9}+-\frac{73}{27}In summary, the minimum value of f(x) is at (-4, 1), and the maximum value is at (2, 1). The value of f(x) at (1, -4) is (-3, a+b+c+d=-5).
- #2
MarkFL
Gold Member
MHB
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Hello, and welcome to MHB! (Wave)
We know two points on the curve, so we may write:
\(\displaystyle f(1)=a(1)^3+b(1)^2+c(1)+d=a+b+c+d=-4\)
\(\displaystyle f(-2)=a(-2)^3+b(-2)^2+c(-2)+d=-8a+4b-2c+d=1\)
Now, we know \(f'(x)=0\) at the two given points as well, so let's first compute:
\(\displaystyle f'(x)=3ax^2+2bx+c\)
Hence:
\(\displaystyle f'(1)=3a(1)^2+2b(1)+c=3a+2b+c=0\)
\(\displaystyle f'(-2)=3a(-2)^2+2b(-2)+c=12a-4b+c=0\)
Now, we have 4 equations in 4 unknowns. Solving this system, we find:
\(\displaystyle (a,b,c,d)=\left(\frac{10}{27},\frac{5}{9},-\frac{20}{9},-\frac{73}{27}\right)\)
Here's a graph of the resulting cubic, showing the turning points:
View attachment 8473
Bob123Bob said:
We know two points on the curve, so we may write:
\(\displaystyle f(1)=a(1)^3+b(1)^2+c(1)+d=a+b+c+d=-4\)
\(\displaystyle f(-2)=a(-2)^3+b(-2)^2+c(-2)+d=-8a+4b-2c+d=1\)
Now, we know \(f'(x)=0\) at the two given points as well, so let's first compute:
\(\displaystyle f'(x)=3ax^2+2bx+c\)
Hence:
\(\displaystyle f'(1)=3a(1)^2+2b(1)+c=3a+2b+c=0\)
\(\displaystyle f'(-2)=3a(-2)^2+2b(-2)+c=12a-4b+c=0\)
Now, we have 4 equations in 4 unknowns. Solving this system, we find:
\(\displaystyle (a,b,c,d)=\left(\frac{10}{27},\frac{5}{9},-\frac{20}{9},-\frac{73}{27}\right)\)
Here's a graph of the resulting cubic, showing the turning points:
View attachment 8473
Attachments
- #3
Bob123Bob
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I currently have those same equations I just don't know how to solve the system, if you don't mind can you show your steps for that or just explain how you did, also thanks for the welcome and reply
- #4
Bob123Bob
- 3
- 0
Im not too familiar with how to solve using matrices, I am pretty sure you have to make the diagonal equal to positive one and then the section below equal to zero but knowing that the answer is all fractions I am not exactly sure how to solve it that way. Although if you could show how to solve it that way I would do my best to learn it and understand properly, but if you could do it just with equations it would be a lot easier for me
- #5
MarkFL
Gold Member
MHB
- 13,288
- 12
Bob123Bob said:
I used a CAS to solve the system, but if I were to do it by hand, let's look at the four equations:
\(\displaystyle a+b+c+d=-4\tag{1}\)
\(\displaystyle -8a+4b-2c+d=1\tag{2}\)
\(\displaystyle 3a+2b+c=0\tag{3}\)
\(\displaystyle 12a-4b+c=0\tag{4}\)
If we subtract (3) from (4) we get:
\(\displaystyle 9a-6b=0\implies 3a=2b\)
If we subtract (2) from (1) we get:
\(\displaystyle 9a-3b+3c=-5\implies 3b+3c=-5\implies c=-\frac{3b+5}{3}\)
Substituting for \(3a\) and \(c\) into (3) we have:
\(\displaystyle 2b+2b-\frac{3b+5}{3}=0\implies b=\frac{5}{9}\implies c=-\frac{20}{9}\implies a=\frac{10}{27}\implies d=-\frac{73}{27}\)
FAQ: Find a,b,c,d given max and min in cubic function
What is a cubic function?
A cubic function is a polynomial function of degree 3. This means that the highest exponent in the equation is 3. The general form of a cubic function is f(x) = ax^3 + bx^2 + cx + d, where a, b, c, and d are constants.
How do I find the maximum and minimum points of a cubic function?
To find the maximum and minimum points of a cubic function, you can use the derivative of the function. Set the derivative equal to 0 and solve for x. The resulting values of x will give you the coordinates of the maximum and minimum points.
Can I use the maximum and minimum points to find the values of a, b, c, and d in a cubic function?
Yes, you can use the maximum and minimum points to find the values of a, b, c, and d in a cubic function. Substitute the coordinates of the points into the general form of a cubic function and solve the resulting system of equations to find the values of a, b, c, and d.
How many maximum and minimum points can a cubic function have?
A cubic function can have a maximum of 2 maximum points and 2 minimum points. This is because a cubic function is a polynomial of degree 3, which means it can have up to 3 real roots. And the maximum and minimum points occur at the turning points of the function, which are the roots of the derivative.
Are there any other methods to find the values of a, b, c, and d in a cubic function given the maximum and minimum points?
Yes, there are other methods to find the values of a, b, c, and d in a cubic function. One method is to use the vertex form of a cubic function, which is f(x) = a(x-h)^3 + k, where (h,k) is the coordinates of the vertex. You can use the coordinates of the maximum or minimum point to find the values of a, h, and k, and then use the resulting equation to find the value of b and c.