Question

PLEASE I'm DESPERATE!
Find a, b, c, and d such that the cubic
f(x) = ax3 + bx2 + cx + d
satisfies the given conditions.
Relative maximum: (3, 11)
Relative minimum: (5, 9)
Inflection point: (4, 10)

Answer 1

I do not have a long period of time to answer this right now, but I will try to help you get on the right track.

There is an inflection point at x=4, y=10. You should make a dot there on your graph, and then put an X on it so you can recognize it.
There is a maximum at x=3, y=11. This means that to the left and right of this point, you are going to be going down. There is a minimum at x=5 y =9, so you will be going up from both sides of this point if you go left or right of it.

You should start with y=x^3+x+2+x and see where that gets you, and in a very short time you will find what your coefficients are by looking at how far off the curves are from where they need to be.

Hope this helps.

Answer 2

take the derivative twice

Answer 3

\[f(x) = ax^3 + bx^2 + cx + d\]
\[f'(x)=3ax^2+2bx+c\]
\[f''(x)=6ax+2b\]

Answer 4

inflexion point is where \(x=4\) so you know
\(f''(4)=0\) which means
\[24a+2b=0\]
hold on to that one

Answer 5

max is at \(x=3\) so you know
\[f'(3)=27a+6b+c=0\]

Answer 6

similarly
\[f'(5)=75a+10b+c=0\]

Answer 7

now we have to solve a system of equations

Answer 8

I thought about that too, but I wasn't sure this student was in calculus. Sounds like a high school algebra problem personally. :)
Hey satellite73, can you check out my problem I posted on existence of a natural number up at the top when you get a chance? :)

Answer 9

i guess you also need that \(f(3)=11\) etc
you will get a system of equations to solve, it is a pain in the butt, but that is what you have to do

Answer 10

I'm so lost. ;A;

Answer 11

you have 4 coefficients to solve for, \(a,b,c,d\)
you need to solve it by finding 4 equations in \(a, b, c, d\)

Answer 12

one we can find by taking the second derivative which is
\[f''(x)=6ax+2b\]
you are told that it has an inflexion point at \(x=4\) right?

Answer 13

this means that \(f''(4)=0\)
so
\[f''(4)=6a\times 4+2b=0\] or
\[24a+2b=0\]
that is one equation

Answer 14

\[f'(x)=3ax^2+2bx+c\] and you are told that there is a local max at \(x=3\)
this tells you \(f'(3)=0\) so
\[f'(3)=3a\times 3^2+2b\times 3+c=0\] or
\[27a+6b+c=0\]there is another equation

Answer 15

now you can find a few more
solve the system of equations and you will get the coefficients