Find a,b,c,d given max and min in cubic function

[Bob123Bob]

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

 

[MarkFL]

Hello, and welcome to MHB! (Wave)

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

 

[Bob123Bob]

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

 

[Bob123Bob]

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

 

[MarkFL]

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

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:

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