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Bob123Bob

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f(x) = ax^3 + bx^2 + cx + d min(1, -4) max(-2, 1) find a,b,c,d

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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).

- #1

Bob123Bob

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f(x) = ax^3 + bx^2 + cx + d min(1, -4) max(-2, 1) find a,b,c,d

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- #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:f(x) = ax^3 + bx^2 + cx + d min(1, -4) max(-2, 1) find a,b,c,d

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

- #3

Bob123Bob

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- #4

Bob123Bob

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- #5

MarkFL

Gold Member

MHB

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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:

\(\displaystyle 2b+2b-\frac{3b+5}{3}=0\implies b=\frac{5}{9}\implies c=-\frac{20}{9}\implies a=\frac{10}{27}\implies d=-\frac{73}{27}\)

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.

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.

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.

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.

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.

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